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.
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?