700 years of secrets of the Sum of Sums (paradoxical harmonic series)

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If I take an arbitrary finite string of digits, say 123456789 and delete all terms of the harmonic series that have denominators containing this string of digits, will the remaining series be convergent?

👍︎︎ 8 👤︎︎ u/Romanmaths 📅︎︎ Nov 27 2020 🗫︎ replies
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Welcome to another Mathologer video today I'd  like to tell you about a bunch of very paradoxical   properties of the so-called harmonic series. This  monster is one of the most iconic infinite objects   in mathematics and has been investigated  for hundreds if not thousands of years.   Of course everybody who knows some calculus  will be familiar with some basic properties   of the harmonic series. However most of its  really amazing paradoxical properties even   many calculus professors don't know.   The reason why I'm telling you about the harmonics  series today are some more counter-intuitive facts   that were only discovered recently and that I  only found out about two months ago. This video   has six chapters each with its own highlights.  Let's have a vote at the very end to figure out   which of these highlights you like best.  Will be interesting and by participating   you'll automatically enter into a draw for  a chance to win a copy of one of my books.   As a bit of a warm-up let's engage in a little  thought experiment. Over there i’ve put weights   of one and two kilos on a simple balance.  Let's assume that the gray bar does not   weigh anything. Remember thought experiment we  can do this :) if i remove the supporting hands   will things stay balanced? Well obviously not. The  red two is heavier so it will go down like that.   So to balance things i have to move  the pointy fulcrum to the right.   But where exactly do i have to put it. Now if  you're familiar with archimedes law of the lever   you shouldn't have any trouble figuring  this out for yourself. Anyway the answer   is one-third from the right. Now here's a nice  two-glance way to see that this is really true.   Put the 2 kilo weight on a little tray like  this. Again we'll assume that the tray does not   weigh anything. Thought experiment remember. Now  split the 2 kilo weight into two 1 kilo weights.   Keep moving them apart up to here. Obviously  the combined center of mass of the two one kilo   weights is still where it used to be. There right  in the middle. On the other hand all the three one   kilo weights are now equally spaced. Can you see  it? There the spacing on the left is two thirds   and the spacing on the right is one third plus  another one third mirrored over is two-thirds.   But then the combined center of mass of the three  equally spaced one kilo weights is right in the   middle that is exactly at the one-third mark from  the right. A very very pretty two glance proof   don't you think? And the whole thing generalizes  nicely? If instead of a two kilo weight we have a   three kilo weight on the right then the balancing  point is one fourth from the right. There   put the tray, spread them out, equally spaced,  right in the middle, works. I like this stuff.   And with a four kilo weight on the right  the balancing point is at one fifth.   Okay when i teach the harmonic series at uni i  usually motivate it with a nice real world puzzle,   versions of which you may encounter in many  science museums the brown blob over there   is supposed to be the edge of a cliff. Now add  20 equal rectangular blocks to this scenario.   What we want to do is to rearrange the blocks into  a stack with as large an overhang as possible.   What do i mean by this? Well i could simply  push the whole stack to the right like this.   This will create this much of an overhang.  There the green bit that's the overhang. Can   we do better? Yes for example we can push  out the one block on top just a little.   There now the green overhang is larger.  Of course we've got to be careful.   Push things over too far and the stack  will topple over. And since the blocks   are not connected to each other the toppling can  happen at any level. For example right up there.   So what's the furthest overhang we can create?  I know hard to tell given only what I've told   you so far. Well so as usual let's look at some  small stacks first to build up some intuition   you know the drill. Right from last time. Now  there a stack with just one block. How far can   we push this block to the right? Easy the center  of mass of this block is right in the middle   and the block will be stable as long as the center  of mass is above the cliff. This means we can push   the block all the way up to here. Now let's say  our blocks are two units long. Then our maximal   overhang is half that. So one unit overhang. Okay  what about stacks consisting of two blocks? Well   with two blocks we can definitely also get one  unit overhang. Right just slide the second block   in like this. There that obviously works, a two  block stack with an overhang of one. But we can   definitely do better because the center of mass of  the whole two block stack in front of us is right   here in the middle. Well because the whole thing  is symmetrical it has to be in the middle right.   And this means that we can slide the whole thing  further over to the right up to this point. Right   the center of mass of the whole stack is above  the cliff edge and so nothing will fall down.   What's the overhang now? Well the one from  before plus one half. And now we just repeat   to get larger and larger overhangs. So  first add a third block in like this.   Next find the center of mass of this new  three block stack. Now unlike in the cases   of just one and two blocks where everything  was nice symmetrical it's not clear where   the center of mass is. However using our warm-up  exercise we can easily figure out its location.   So we can think of our three block  stack as consisting of two parts.   The new block highlighted in blue and the old  two blocks stack. Now the center of mass of the   two blocks stack is already part of the picture.  The center of mass of the blue one is here right   in the middle. How far apart are the centers? Well  that's half the length of the blue block so that's   one. Next our two stack weighs twice as much as  the blue block and this means we're dealing with   exactly our warm-up setup. And this means that the  combined center of mass of the three block stack   is one-third in from the cliff  edge. Ah there nice animation   nice. Now push to the right like this and  we get an overhang of one plus one half plus   one third. Repeat again there slide in, find  the center of mass, nice animation again, and   yes slide over and at this point the overhang is  one plus one half plus one third plus one fourth   and we now know exactly how this will continue.  However also notice that at this point   the top block is completely beyond the cliff edge.   And this is pretty amazing isn't it? A challenge  for you: actually perform this experiment. So use   identical brick-like objects like books to build  a stack at the edge of a table in which the top   block completely clears the table. Link to a photo  of your little stacking miracle in the comments.   Pretty tough challenge. Anyway rinse and repeat  and we get this remarkable leaning tower.   And the overhang we can achieve this way with  20 blocks is? Well one plus one half plus one   third all the way to one twentieth which is  approximately three point six. Very cool.   What I've just shown you is a Mathologerization  of a construction that's at least   170 years old. These days the stack over there is  often referred to as the leaning tower of lire.   This name was coined by the physicist Paul  Johnson and was the title of a note by him   that appeared in the American Journal of  Physics in 1955. There that's the article   the lire in the title was the name of the Italian  currency prior to introduction of the euro and in   his paper Johnson actually plays with a stack  of coins rather than a stack of bricks. Anyway   now a very natural question to ask is how much of  an overhang can we produce if we have an unlimited   supply of blocks. To answer this question we  have to evaluate the sum of the harmonic series.   There she is, the lead character in  today's tale of mathematical mystery.   I'll puzzle out the sum together with you in a  moment but not before i say a little more about   our original puzzle. To find a maximal overhang  by stacking these 20 blocks over there. I'm fairly   sure that 99% of all the people who are familiar  with this leaning tower are under the impression   that this tower produces the largest possible  overhang using 20 blocks. This is not the case,   not even close. Want to see what the optimal  stack really looks like? Ready to be disgusted?   Here's the optimal stack which was only  discovered something like 10 years ago.   The same 20 blocks but reaches out  much further. And ugly as hell right?   And didn't they always tell you that all  optimal maths is automatically beautiful?   While you often encounter beauty  in the optimal this is definitely   not always the case? But of course beauty is in  the eye of the beholder and there's definitely a   lot of ingenuity to be admired here. For example  the use of counterweights and bridging weights   to weigh down pieces that would otherwise  fall down. What the presence of these   bridging pieces also implies is that you cannot  actually build this weird stack layer by layer   from the ground up. Right, building up to here  the piece in front would definitely fall down.   Now whenever i talk about this someone will ask  whether it is at least true that the leaning tower   is the optimal stack if we allow only one brick  per layer. And this is actually true. Although a   correct proof that this is true for all possible  numbers of bricks was only nailed down in 2018   by David Treeby somebody from Monash, in an  article in the American Mathematical Monthly.   That this maximal overhang property is not  completely obvious is hinted at by the fact   that there is actually another one block per  layer stack that has exactly the same overhang   as our leaning tower of lire. Again hardly anybody  seems to be aware of this. Have another look at   the two top blocks. Can you see where I'm going  with this? Center of mass right in the middle.   Well simply flip them like this and  you get that second maximal stack.   Cute right? If you're interested in  chasing down this particular rabbit   hole I've linked to some nice links  in the description to get you started.   So we can arrange for one complete brick to  be beyond the cliff edge using four bricks.   How can you do the same with only three  bricks? Give you solutions in the comments.   Okay so we would like to know how much of an  overhang one of our towers of lire can have if   we have an unlimited supply of blocks. To answer  this question we have to figure out what the sum   of the harmonic series is. Well let's start  adding. Okay so we are adding more and more   positive terms and so obviously these partial sums  will get bigger and bigger. This means there are   really only two possibilities. One, these partial  sums converge to a finite number which we would   then declare to be the sum of the series. Or  the partial sums explode to positive infinity.   Which one is it? Obviously many of you will know  the answer but imagine for the moment that this is   the first time you see the series. What would  you bet your life on? Finite or infinite sum?   Well the terms we're adding get smaller and  smaller and amount to closer and closer to   nothing, so shouldn't the partial sum settle  down to something finite, just like for some   other famous subseries of the harmonic series,  like this one here. What's the sum of this one?   Well pretty obvious… so there, infinitely many  positive terms adding up to something finite.   What about the sum we're interested in.  Well let's not drag it out any further.   Turns out the harmonic series explodes to  infinity and the first in my opinion nicest   proof of this fact goes back almost 700 years  and is due to the french bishop Nicole Oresme.   Hmm clearly some posture problems due  to excessive obsessing with mathematics.   Okay here's this proof animated  as best as i can. Enjoy.   An absolute gem of a proof  don't you think? Of course   what this implies is the leaning tower  paradox, that at least in theory we   can make our overhang as large as we  wish by adding more and more blocks.   Marvelous stuff isn't it? There are  actually a couple of other spectacular   real world paradoxes that are based on the  fact that the harmonic series is exploding.   For example if you have not heard  of the worm on a rubber band paradox   definitely check out the links in the description  of this video. And if you know of any off the   beaten track real world harmonic paradox please  share it with the rest of us in the comments. Okay   what have we got so far? One glance balancing,  leaning tower of lire, bizarre maximal overhang   towers, and a 700 year old proof by a  bishop. What do you like best so far?   Okay so the harmonic series diverges to infinity,  but it does so extremely slowly. Have a look.   As you can see the series really grows very  slowly. That's a million terms we've added here.   How many terms do you think it takes  for the partial sum to hit 100.   That number was calculated for the first  time in 1968 by the mathematician John w   Wrench jr which was a bit of an achievement  at the time. Here it is. So to get to 100   you have to add in the order of 10 to the 43  terms. That's really mind-boggling isn't it?   Also how did John Wrench figure out that  exact number? Obviously actually adding   the first 10 to the 43 terms one at a time is  completely out of the question. Even the most   powerful computers would take gazillions of  years to add up that many terms one by one.   Well maybe there's some kind of magical formula  for the n's partial sum that allows us to solve   for the answer? Such a formula would also allow  us to figure out that all important question how   much of an overhang a leaning tower of one google  blocks that is 10 to the 100 blocks will produce.   Well maybe not that important  question itself but formulas like this   really are incredibly important. Anyway  let's hunt for such a magical formula.   If you watched the last what comes  next video you know what comes next.   We start summing again from the beginning but  this time write down the partial sums as fractions   let's see whether we can guess a  pattern that translates into a formula.   Okay here comes the dreaded question: can you see  a pattern? Well looks like a complete mess doesn't   it? But there's something striking about all these  fractions. Can you see it? No okay here's a hint.   What do all the numerators have in common and  what do all the denominators have in common?   Yeah of course many of you will have spotted  straight away that all the numerators are odd   and all the denominators are even. Coincidence?  I don't think so! In fact if you keep on going   you keep getting partial sums that  are odd divided by even fractions.   Interesting but so what? Well actually in terms  of odd and even there are four different types   of fractions. Odd over even, even over  odd, even over even and odd over odd.   There's one thing that is special about odd  over even fractions and that is … well, think   integers … if you divide an odd number by an even  number the result can never be an integer. Right?   Whereas with the other types of fractions  they can actually be integers in disguise.   There all those fractions are really integers  in disguise. Nice little insight isn't it?   An odd divided by an even  fraction cannot be an integer.   That could be a starting point for a nice  number magic trick. Have to ponder this.   Anyway once you notice this odd over even pattern  in the sequence of partial sums it's actually   not hard to prove rigorously that in fact all  infinitely many partial sums are of this form,   with the exception of the first partial  sum 1 of course which is an integer. And   this then also shows that apart from that  1 at the beginning there are no integers   among the partial sums. That's a pretty  amazing result when you think about it.   Right? So these partial sums creep up slower  and slower smaller in smaller increments.   On the way to infinity they pass every  single one of the infinitely many integers   and manage to miss them all by my minuter and  minuter amounts. Amazing isn't it? But in fact   something even more remarkable is true. Add up a  randomly picked couple of consecutive terms of the   harmonic series like these ones here. Consecutive  is the important bit here. Then it can be shown   that all of these sums combine into odd over  even fractions and are therefore never integers.   The first proof of this fact was published in  1918 and is due to the Hungarian mathematician   Joseph Kurschak the very friendly looking guy  on the white chair in the picture. That's a very   typical mathematician group photo 100 years ago:  all men, ties and suits and no t-shirts. Times   have definitely changed. Okay to finish off this  chapter here is another challenge question for   you: can you write the numbers 2 and 3 as sum of  reciprocals of distinct positive integers. Hmm i   guess i got a bit sidetracked there. Still i think  this detour was totally worth it and this is also   a nice example of how scientific discovery works.  You don't always find what you're looking for.   Anyway remember what we set out to find at the  beginning of the last chapter was an easy formula   for the n's partial sum of the harmonic series.  The usual approach to look at a couple of small   examples to try and spot a pattern that amounts to  a formula actually does not work for the harmonic   series but there is another very nice and natural  way to approach this problem. One, one half, one   third, one fourth 1/n s. Doesn't this cry out for  us to ponder our sum in terms of the function 1/x.   Well let's plot it. Where are the terms of our  sum hiding in this picture? Well plug in one,   one over one is one so there is one.  Obviously that blue segment is one unit long   but there's another one hiding here, a one unit  area. Can you spot it? There that's square.   One half is here, and as an area here.  One half times one is one half and so on.   And so there that's a simple visual area  counterpart to the harmonic series. Of course   there is another area in this picture that cries  out to be highlighted that area under the curve.   But how is that going to help? Well we're  chasing the partial sums of the harmonic series.   Let's focus on a small example. 1 plus 1  half plus one-third plus one-fourth is equal   to the blue area. Turns out that there is a really  easy formula for the corresponding area under the   curve. Integral baby calculus 101 a lot of you  would have done this in school tells us that this   area is exactly equal to the natural logarithm of  5. And of course that means that our partial sum   is approximately equal to that logarithm. In  general this approximation looks like this.   How much of a difference is there between the left  and right sides? Well that difference is the area   of the remaining blue bits sticking out on top  and luckily these blue bits don't add up too much.   Let me show you. See that one times one square  over there? Well all the blue bits fit in there   and so the sum of the blue bits, the difference  up there is less than one. And that's important no   matter how humongous the partial sum is. In fact  this tethering of the logarithm and the partial   sum immediately implies another proof of the fact  that the harmonic series explodes to infinity.   Right, since the logarithm explodes to infinity  the tethered partial sum gets dragged along   to infinity doesn't have a choice at all in  this matter. Anyway since for large partial   sums an extra one doesn't really matter, for  all intents and purposes the n's partial sum   is equal to the natural logarithm of n plus one.  Having said that we can still improve this simple   very good approximate formula  dramatically with very little effort.   Notice that really all infinitely many blue bits  fit into the square on the left. This means that   all the blue bits add up to a number less than one  this number has a special name it's called gamma,   or the Euler-Mascheroni constant or simply Euler's  constant. Euler again can't escape him right.   Another challenge for you: okay so gamma is less  than one. That’s obvious right. But can you also   see at a glance that gamma has to be greater than  0.5. If you can, tell us why this is obvious.   Anyway Euler was the first to investigate  gamma around 1734 and calculated the first five   decimals. Gamma is one of those mathematical super  constant that just like pi, phi and e pops up all   over the place. Just as a little teaser here is  one of my favourite identities involving gamma   they are e gamma and all the prime  numbers in one formula. How crazy is that?   Anyway here is a super nice trick let's add  gamma to our approximate formula for the   partial sums of the harmonic series. Here's what  the sum corresponds to in the picture so the sum   is exactly equal to the pink partial sum one  plus one half plus one third plus one fourth,   plus a tiny portion of gamma corresponding to the  extra blue bits petering off to the right.when   you think about this this is really an amazing  formula. As n and with it the partial sum gets   huger and huger this formula approximates  the partial sum better and better. That's   the exact opposite to what most formulas  approximating exploding sums usually do.   Just two examples to show you how amazing this  formula really is for n is equal to 5 we get this.   So in terms of correct digits the formula  gets the integer part of the partial sum   right but no decimals. Now here's what you  get for n is equal to five hundred thousand.   That's already five correct decimals and it  really gets better and better the larger n is.   Absolutely marvellous isn't it? To finish this  chapter let me tackle those two questions that we   set out to answer using our formula. Remember the  humongous exact number of harmonic terms you have   to add to get a sum exceeding 100?  That number John Wrench calculated.   What does our formula suggest this number is? Well  we want 100. Let's solve for it. There autopilot…   and well if you actually evaluate the expression  on the right for example in Mathematica you arrive   at a number that's just shy of John Wrench's exact  number. According to Mathematica the difference   is about 0.63. Pretty good. Our gamma is really a  very magical constant. In fact John Wrench's proof   that his number is it is based on our formula,  starts with our formula. Of course we can also use   our formula to figure out how much of an overhang  a leaning tower with n blocks can produce.   So how about one google blocks for example. Well  the natural logarithm of 10 to the 100 is 230   and a bit units this thing really grows very  slowly. Well that's logarithmic growth for you.   In comparison it has been shown that  there exists a positive number c   such that by using crazy towers an overhang of c  times cube root of n is achievable for large n.   That's exponentially better than logarithmic  growth. Nobody knows exactly what the absolute   best crazy tower looks like for large numbers of  blocks but one not so crazy family whose overhang   grows like cube root of n are these parabolic  towers that were discovered by mike paterson   and Uri Zwick around 2009. The example over there  consists of 571 blocks and produces an overhang of   10 units the smallest leaning tower that has  this overhang contains 12 367 blocks. Good stuff   and there are still lots of nice open problems  that need to be sorted out. Maybe by you. So   to finish this chapter let me at least mention  one more super interesting fact about gamma.   Since you are watching this you're  probably all aware of the fact that there   is a mysterious and important connection  between the sum of the positive integers   one plus two plus three and the number minus  one-twelfths. Despite what some people claim   this wonderful relationship is not an equality.  Now when we dig deeper we find that in a very   precise sense -1/12 is to the sum of the positive  integers what gamma is to the harmonic series,   the sum of the reciprocals of the positive  integers. I don't have any time to elaborate   on this today. If you'd like to do a bit of  research on your own google Ramanujan summation   and find out why the Ramanujan sum of one plus  two plus three -1/12 and the Ramanujan sum of the   harmonic series is gamma. Okay what have we got so  far: one glance balancing leaning, tower of lire,   bizarre maximal overhang towers, a 700 year old  proof by a bishop, no integers among the partial   sums, magical approximate formula and of course  gamma. And i even managed to squeeze -1/12s in.   What do you like best so far? Remember  there will be a vote at the end.   Okay final chapter: to finish off let me just tell  you about that other remarkable harmonic fact that   i only found out about two months ago and that  inspired me to dedicate this video to my good   old friend the harmonic series. The harmonic  series is the sum of sums in the sense that   it contains many many many many very important  infinite series: for example there's the geometric   series of the powers of one half there which sums  the two. Then there's Euler's super famous sum of   the reciprocals of the squares, the reciprocals  of the fourth powers and in general all those   other reciprocals of integer powers series  that are part of the Riemann zeta function.   Then there is e as the sum of the reciprocals of  the factorials and so on and so forth. I've talked   about these and many other famous subseries  of the harmonic series in previous videos.   Now obviously there are infinitely many sub-series  contained in the harmonic series summing to   infinitely many numbers and so a natural question  to ask is whether every number is such a sum.   What do you think? What's your gut feeling? The  answer is yes every positive number occurs as a   sum of infinitely many sub-series of the harmonic  series. This is actually very very easy to see?   Let me quickly show you how you can find a  sub-series that sums to our new favourite magic   number gamma using a greedy algorithm. Works for  any other positive number. As the first term of   a subseries that will sum to gamma choose the  largest harmonic reciprocal less than gamma.   Well gamma is 0.57 and so the largest  harmonic reciprocal less than gamma is   one-half. The next term of our  sub-series is the largest harmonic term   that when added to one-half gives the sum  less than gamma. That happens to be 1/13.   There that sum is just a little bit less than  gamma and you just keep on going like this, always   add the largest possible harmonic term that gives  you a sum less than gamma. This will automatically   generate a subseries containing infinitely  many terms that adds very rapidly to gamma.nice   and super simple this is an example of a  greedy algorithm in action. We always greedily   go for the largest possible reciprocal  that fits. Of course apart from all those   sub-series with finite sums there are also plenty  of sub-series that just like the harmonic series   diverge to infinity. For example the sum of  the reciprocals of the even numbers diverges.   Pretty obvious because the series is also just  the original series, there's the original series   again, multiplied by one half. So the sum of the  reciprocals of the even numbers is one half times   infinity which is still infinity. What about the  sum of the reciprocals of the odd numbers? Well   the first odd reciprocal one is greater  than the first even reciprocal of one half.   The second odd reciprocal one third is greater  than the second even reciprocal 1/4 and so on.   And so the sum of the odd reciprocals is  greater term by term than the infinite sum   of the even reciprocals and so also has to be  infinite. Okay maybe that's not that terribly   surprising. Here's something trickier. How  about the sum of the reciprocals of the primes?   Finite or infinite? Well what do you think? Of  course there are infinitely many primes but they   are pretty sparse with arbitrarily large gaps in  between them. Does this mean the sum is finite?   Well those of you who watched my video on Euler's  product formula for the Riemann zeta function   may remember that we proved,  following Euler's footsteps of course,   that the surprising answer is this sum is  infinite. Really not obvious at all. Let's   see how much of an intuition you've developed  for these infinite sums. I'll show you a couple   of sub-series and you try to guess whether  they have a finite or infinite sum. Ready?   Okay that's the full harmonic series over  there. First let's skip every tenth term   starting with one ninth. That takes  out this infinite column of terms.   Is the leftover series finite or infinite?  You got it? Easy right? This sub series must   explode because, well, there are lots of  ways to see this at a glance. For example   it must explode because it contains all the even  reciprocals which we've already seen do explode.   For my second mystery series i’m going  to throw away all the reciprocals of   numbers that contain the digit nine. Okay  so that gets rid of the last column again.   Right all those have a nine. But of course there  are numbers other than these that feature the   digit nine. For example all the guys at the bottom  there. Okay no more nines as digits left over.   Finite or infinite? Imagine your life depends on  getting this right. You've got five seconds five   four three two one zero. You got it? Well based on  past experience asking this question almost all of   you will have gone for infinite. Which is wrong.  Believe it or not but this one has a finite sum.   I first encountered this no 9th sum in the  second week at uni in Germany a couple of   ice ages ago. They actually gave it to us as  a homework exercise. This fun series is known   as Kempner’s series named after the mathematician  Kempner who published a finiteness proof in 1914.   What makes this counter-intuitive results  particularly memorable for me is that i   did come with a finiteness proof myself  as part of this homework assignment.   I can't remember how i proved it but i do remember  that i got zero marks because the marker could not   be bothered to actually engage with my proof that  was different from what was expected. Grrrr :(   I'll do a super nice animated algebra proof  that the no 9 series has a finite sum at the   end of this video but just real quick one way  to intuitively see that Kempner's no 9 series   may be sparse enough to allow for it to converge  is to realize that the picture over there is a   bit misleading, that in fact the vast majority  of integers with large numbers of digits will   contain a nine among their digits and therefore  the reciprocals will not count towards our sum.   Tristan one of the three volunteers who helped  me with proofreading and polishing my slideshows   suggested a nice visual way to illustrate  this thinning out. Make the 10 x 10 picture   over there a bit more squarish and color the  part with the remaining reciprocals black.   Let's pretend the upper left square is also  black. Makes for a cleaner argument. Okay   then up to that little fudging in the corner  the black area inside the big square tells   us how many of the first 10 times 10 terms of  the harmonic series are contained in Kempner's   no 9s series. Zooming out the corresponding  picture for the first 100 times 100 that's the   first 10000 terms of the harmonic series looks  like. There is more white than before and so   the percentage of omitted terms has gone up.  Interesting. Here's the picture for one thousand   times one thousand that's the first one million  terms. Of the harmonic series the percentage of   omitted terms has gone up again, now ten thousand  times ten thousand. Looks the same but it's not.   On closer inspection there are actually tiny  white subdivisions of the little black squares   that mirror the nine times nine subdivision  that you're seeing in front of you. Like for   example the little square right there. When you  have a close look it actually looks like this.   Because of the presence of the white subdivisions  the little square is actually gray and not   completely black and so again the percentage of  white has gone up. As we zoom in digit by digit   the subdivisioning continues turning what you  see in front of you into a fractal-like object.   And the percentage of white will go to  100 percent as we push things to infinity.   Pretty amazing. Thanks again to Tristan for  suggesting this fun visualization of the really   dramatic thinning out of the harmonic series when  we throw away all the reciprocals of integers that   contain a digit nine. Makes the convergence of  the series so much more believable. Not a proof   but as i said I'll do an animated algebra proof at  the end. Okay everything I've told you so far I've   known for many many years now. What i stumbled  across two months ago were a couple of articles   by the mathematician and computer scientist  Robert Baillie full of very interesting and   surprising results about the no 9 series and a  whole universe of closely related crazy sub-series   of the harmonic series that i’d never heard of  before. For example it turns out that nobody   actually knew the sum of the no 9 series until  quite recently. What Kempner showed in his 1914   article was that the sum of the no 9 series is  less than 90. How close is 90 to the actual sum.   Well although the no 9 series turned into being  a bit of a celebrity among people in the know,   nobody appears to have given this question  any serious thought, until Robert Baillie   did so about 30 years ago. He figured out that  the sum of the no 9 series is approximately   22.92. At that point in time he also figured out  good approximations of the no zeros, no ones, no   twos etc series. Here's a table from his article  which summarizes his results. Why had nobody   calculated any reasonable approximations before?  Well it turns out that all these missing digits   series converge at a snail's pace. For example for  the no 9 series the sum of the first 10 to the 28   terms is still less than 22, so not even close  to what most of us would call a reasonable   approximation of the true value of the sum.  And so for these series it's also completely   unfeasible to simply keep adding up terms until  we get close to the true value. You really need   a good idea to get anywhere with approximating  these sums. More recently Robert Baillie also   managed to calculate similarly good approximations  for other closely related crazy series like the   no zeros and no nine series (so a couple of digits  missing not just one), the at most 10 9s series   (so every term has at most 10 nines), the exactly  100 0s series, and so on. Really impressive stuff   especially since most of these series converge  much much much slower than the already very   slowly converging no 9 series. For example the  sum of the 100 0s series is approximately 23.03.   Also pretty amazing in another way when you think  about it right? The first term of the 100 series   is one over one google which of course is a  crazy small number and yet the sum of the series   in which every term is exactly 100 zeros ends up  being larger than the sum of the no 9 series which   starts out with much larger terms. So how are  these really good approximation calculated? Well   it's not that terribly complicated but  also quite nitty gritty technical so i   won't go into details here. The basic idea  is to exploit the patterns in these series   in the calculations, the kind of patterns that  were visible in Tristan's fractal for example.   For the keen among you I've included some  relevant links in the description of this video.   Anyway as you can probably tell Robert Baillie's  papers really made my day when i stumble across   them and i hope I've been able to convey to  you some of my excitement about these new   and old insights into the nature of the harmonic  series and infinite series in general. What i’ll   finish with today is a Mathologerized slightly  refined version of Kempner's original proof   which shows that the son, sum not the son  of the no 9 series is less than 80. Watch   it and afterwards cast your vote for the harmonic  marvel in this video that you enjoyed most. What's   most memorable: one glance balancing, leaning  tower of lire, bizarre maximal overhang towers,   that 700 year old proof, no integers among the  partial sums, amazing approximate formula, gamma,   greedy algorithm, Tristan's fractal, finiteness  proof. Well we really did cover a lot of ground.   Cast your vote by leaving a comment within  the first two weeks after this video   went live for a chance to win one of  my books. I'll then pick a random name   among those who participated and a random  book and will I’ll announce the winner   as part of the 2020 Christmas video. Okay now to  finish here's why Kempner’s no 9e sum is finite.
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Channel: Mathologer
Views: 322,407
Rating: 4.9512644 out of 5
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Length: 46min 34sec (2794 seconds)
Published: Sat Nov 21 2020
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