A lot of you will be familiar with this
strange identity here: 1+2+3+ ... is equal to -1/12. Now when you
think about it this looks totally insane because what we're doing here on the
left side is we are adding larger and larger positive values, infinitely many of them,
so if anything this left should add up to plus infinity. And not to negative 1/12.
Most people who look at this for the first time think that this is just nonsense. Except it comes up in a very very famous letter in mathematics. A letter
that was written by this guy down there Srinivasa Ramanujan in 1913. It was addressed to G.H. Hardy in Cambridge, UK. one of the most famous mathematicians at the time. Now a lot of people have heard about this letter but have not seen it. So I thought I
show it to you. Ok, so here is the letter. It is a long letter and it's basically lots and lots and lots of theorems, mathematical theorems, enumeration of lots of theorems that
Ramanujan says he's found and the most remarkable thing about this is that
he has no real former mathematical training unlike Hardy. And so a lot of
these things are pretty amazing and towards the end of the letter well this strange identity here comes up,
and a couple of similarly strange identities around it. And actually this
letter got an invitation to England to work with Hardy. So, you know, maybe we
should have a really close look at this and that's actually what I want to do today. Now,
I'm not the first one to do this Numberphile's done a couple of videos on this,
and other people, but I think I'd like to really try to do this properly, so it's
going to be a very long video and so to make it accessible to as many
people as possible I've split into four parts, so four levels
of enlightenment, and you can pretty much stop anywhere and, you know, get something
out of it this way. So let me know how far you get with this video. Since, eh,
Ramanujan was a devout Hindu, I thought I'll ask the elephant god to help us out here, to
lead us through these four different levels of enlightenment when it comes to
these strange sums. Ok, here is level 1: Just do it. Now, Ramanujan actually doesn't tell how
he derives this strange value in his letter to Hardy. You know, it's
just there, he just says "that's what I get". Now, we have his notebooks, and there is
one of the pages from the notebook where he actually talks about this. In the
previous page he also talks about this; it is more like an afterthought to what he
does on the other page; but this one I can actually do in a video like this so
keep it completely elementary I can do this one here. Let's have a closer look at what he
does Ok, so he starts by saying C. And he doesn't call it a sum he says it's a constant C is equal to this strange sum. And then he
does a couple of logical deductions, and at the end of that he gets to -1/12. Not much, so should be able to do this right? Ok, so here we go. So, he says there's
something there. That something - I call it C. C is equal to this sum. Now, if that
something is halfway as reasonable, I should be able to manipulate this as usual.
One of the things I should be able to do is multiply it by 4. So he does it. So,
4 times C is equal to... 4 times 1 is 4. I put the 4 under the 1. No I put it under the 2. Whatever Let's just go with it. Then 4 times 2 is 8. Again I skip one. Put the 8. Then I always skip one and put down all the multiples of 4 here in the
second line and now the next step is to subtract the bottom from the top. we get C minus 4C is -3C. 1-0 is 1. 2-4=-2. 3-0=3. 4-8=-4. And so on So we get this nice sum here 1-2+3-... etc Now comes something surprising. Now, Ramanujan somehow sees that this is equal to a 1/(1
+1) squared. Well that's 1/4. So, if that somehow makes sense, well, then we can say
-3C = 1/4 Just solve for C, and we get -1/12 Ok, we still need that step here, that's not
clear at all. So how do we get that? That's actually has two ingredients.
The first ingredient is that Ramanujan knows that (1-2+3-... etc) sum is equal to the (1-1+1-... etc) sum squared Ok... Second ingredient is this here: that the
(1-1+1-... etc) sum is equal to 1/(1+1) Let's try to make sense of
this. First one. So, here we've got an infinite sum that's multiplied by itself So, how do we do this? We have to multiply every term at the top by every term at the bottom and then add up
everything. So, let's arrange this like this so this is the first sum here that's
the second sum here. Now we do first term times first term, that's 1 times
1 is 1 Now first term times second term is (-1) first term times third term is 1 and so on And that actually gives us this nice checkerboard
pattern of plus and minus ones How do you sum up this checkerboard pattern here? Well, you do it in diagonals. So, here we go. We're gonna put in diagonals. And now what's the sum we're actually forming here? Well, blue diagonal is 1, green diagonal is -2, yellow diagonal is +3, next diagonal is -4 and so on. So this then clarifies what
what he means by this What about the second step? that the (1-1+1-... etc) sum is equal to 1/(1+1) Well, that comes from something that a lot of you will be familiar with: this identity here Sum of the geometric series. Now, this is not valid for all R. It's only valid if the R is kinda small. If it's in the range from -1 to +1, this actually works So, for example, if you choose R is equal to 1/2, then
this whole thing turns into 1+1/2+1/4+1/8+... and so on is equal to two. So, you know, a lot of
people would have seen this. A lot of people will be familiar with this. As I said, this works in the range from -1 to 1. It doesn't work for -1, it doesn't work for 1. At least not in standard calculus But what Ramanujan now does
(and a couple of people before him actually) is substitute R = -1 anyway and
when you do that what you get: Here you get -1, Here you get 1, Here you get -1 and so on So that gives you this identity Then, of course, if you substitute
this into that, you got exactly what Ramanujan wants. If you know any
calculus, you think like every single step here is totally insane. So there is the sums here, right.
This one here, that one here, that one here We're subtracting infinity from infinity, so nothing here makes sense. So I've got my insaneometer here at the left. Let's give this a score of 4 on
the insaneometer. Well, insane, yes, but remember: this guy is a genius and this
guy here G.H.Hardy, well, he's... he's no dummy either. So
somehow they can take these things seriously, so we better have a close look. Okay, so what does Ramanujan do here? Well, he actually does something that a
lot of mathematicians do when they don't know what to do. So, often, you know, you're
encountering expressions like this is somebody that's kind of scribbled down on a
piece of paper something with lots of dots in it, and to
start with, that's just a couple of symbols on a piece of paper and you don't really
know what they mean, right? So don't really know what they mean. You know, if there's just
finitely many (without the dots), usually know what it means but if there's dots in
there, you know, it gets tricky You don't know what this means, right. Now, you can try
and kind of just straight away develop some theory that makes sense of
these sorts of things, but often it's a better idea just kind of do it. Okay? So what
you do is you don't know what these things are so you just let them be some
sort of unknown value, and then you just kind of go for it. So if there's
anything reasonable, you should be doing this sort of manipulations that maybe are not
too wild and then maybe get you somewhere. And let me just demonstrate this. This is
kind of a very famous one. You've got this infinite sum (that is actually a geometric series that we just looked at), and... What can I do here? Well, if that's anything reasonable, I should
be able to multiply (just like what Ramanujan does β at the beginning he multiplies by 4)
I multiply by R Ok? So I multiply R times C, then on the other side
we get 1 times R (put R here), then R times R β R^2 (I put it there), and so on. And that will give me
this; and I do exactly what Ramanujan does and subtract the bottom from the
top. So C minus RC - there we go; then here, well, here we are really lucky, a lot luckier
than Ramanujan was with his sum. Everything here pretty much cancels out, and the
only thing that's left over is the 1. And then we can for C, and we get this. Ok. And so,
what we've found here is that assuming that this somehow makes sense and that we can manipulate it as usual, [we] necessarily get that this has to be equal to 1/(1-R). Now, later on
in calculus, when we actually really go for it we actually say "well, this does make
sense when R is in this range, but only then". Yeah? So for other values of R, like -1, that one here, that doesn't make sense;
that gets discarded, that's evil, we don't want to know this, we don't touch this in standard calculus. But actually it turns out that, you know, if you just setup mathematics in
a slightly different way, this might actually make total sense. So on a different
planet that might be totally okay. I could also try R=2, for example, in here,
and then you get something really strange, which is very close to the sum
that we actually started with; it's 1+2+2^2+ +2^3... that's all positive values getting bigger and bigger, and should add up to infinity, but this guy here says it's actually -1. So how does that work? Well, it turns out that if you're just on the right
sort of planet, there is mathematics that makes perfect sense and in which this is actually true. Ok, and now, that was level 1, so you survived so far? Perfect. Now, if you want to actually know how we make sense of all these things stick around for Level 2 :) Ok, so in Level 2 we're actually going to focus not so much on what happens on other planets but what
happens on Earth. So Level 2 is about following the rules. What are the rules for dealing
with these sorts of things in standard calculus? What makes this right and
all of these wrong? Well, short answer is that this guy here on the left, which is
called an infinite series, is convergent and has sum 2, whereas all of those
guys here are divergent. Ok, let's have a close look. Now, at some point in
time, somebody wrote down this infinite sum, and at that point in time, actually,
nobody knew what it really means. Well, you know what (1+1/2) is, you know what (1+1/2+1/4) is. But what does it mean to add up infinitely
many things? To start with, you know, there's no... that doesn't have any meaning. Alright, but you can try to give it to some meaning. So we'll just start adding, ok? So we've got 1
(put it down); we add 1/2 to 1 - we get three halves, and then we add 1/4 to
get 7/4. We just keep on going like this and we get a sequence of numbers
down here that goes on forever. And when we have a close look at this
sequence of numbers, the sequence of partial sums, then we see... Well, they get
closer and closer to 2. And actually 2 is greater than all of them. And in fact,
2 is the smallest number greater than all of these fellows, all of these partial sums. And if you look at this expression, well, if
it's to mean anything, really the most natural thing to say here is that this should
be equal to 2. And actually it's a definition, it's, you know, a definition
that mathematicians have made. It's our choice to say that this something is
equal to 2, it's a definition. It's a very natural one, it's probably the most natural one,
but it's still a choice. Our choice to do it this way and not another way. In fact, there is
another human choice in there, and (human choices will always gonna be in the background whenever you deal with these sorts of things) it is that we're actually always, by default,
talking about real numbers. Okay? So we're talking about real numbers; that's how
we make sense of things. There would be other possible choices for the default
number system that we use, and on different planets, maybe, they do something
different; for example, the surreal numbers or the hyperreal numbers. But for everything
that we do here on earth as a default it's the real numbers. It's also important to keep in mind. Now, why was our choice of definition so good? Well, because, to a large extent, you can actually
manipulate these sums just like you would manipulate finite sums. You know,
this, you know, sometimes not, but... For example, if you've got a convergent
series here and its sum is A, then you can actually multiply the right side here
by something, for example 5, and you get something convergent again. The series
that we get down here is convergent again. And it converges to... has the sum 5*A, as
expected. So that's really good. Or, if you've got two of those guys, right, so infinite
series (sum A), infinite series (sum B), we can term-wise add things, and that gives us
another infinite series that converges. And the sum is A plus B β perfect! Or we can
subtract the bottom from the top and it would be also, I think, A minus
B β perfect! So what that means is in particular, is that if this guy here is a
convergent sum, then all of these manipulations that with did here are perfectly
OK, and we get the right result. So we multiply, we subtract, it works again,
and we get the right result. Now, in standard calculus courses you actually
usually don't get to this point, as actually some bits about our definition that are not so
great. So, for example, if you've got something that works β convergent,
convergent β and you multiply them together, you would expect that you
always get like A times B, something convergent with the sum A times B,
and often that is the case, but sometimes it actually isn't. So, for example, this guy here, that's convergent and has a sum, but when you actually do to the square of this,
this sum by itself, that doesn't converge. And it's a bit of a problem. Let's
just keep this in mind OK. So, this is maths on Earth. So let's
just have a look at these sums here. Why don't they work? Well, let's have at the first one here. Partial sums, we need partial sums. So first
partial sum is 1. 1-1 is 0, +1 is 1, -1 is 0, and so on, so 1, 0, 1, 0, 1, 0. That doesn't
converge to anything, that doesn't have a limit. So it's, you know, it doesn't have a
sum in the usual sense. So that means it's divergent. What about the second one? 1, 1 minus 2 is -1, plus 3 is 2, minus 4 is -2, and so on. Also doesn't settle down β forget about it. Last one. 1, 3, 6, 10 β kind of explodes. All these series here, on the left side, are divergent.
They don't have a sum. Definitely not 1/2 here, definitely not 1/4 here,
definitely not -1/12 there. Hm, but when you kind of look at it... These sequences that we get to, the
sequence of partial sums You know, they're very different, so instead of
calculus, you could just take everything that's not convergent and just say "well,
throw it away". So you don't really look at the difference that you get here, this,
you know, graduation. There's actually different sorts of divergence here, different sorts of divergence. And this one is
kind of a tame one, this is... Well, Well, this is kind of oscillating around
a common center here somehow. And this kind of explodes. You know, maybe you want to capture the
differences between these different sorts of divergence. So let's have a
look at the first one here. So it's a very very famous one; actually, the first thing
you see in calculus pretty much saying that, you know, this doesn't work,
you know, for various reasons. Now, partial sums: 1, 0, 1, 0... This is supposed to be equal to 1/2. Well, how to get one half out of 1and 0? Well, somehow do the average. So average of 1 and 0 is 1/2. And actually
there's a very neat way of defining the sum of one of these
series in a different way. So what we do is: we don't stop with the sequence of
partial sums, but now what we do is we generate a second sequence of averages.
So, average of 1 is, well, 1; average of 1 and 0 is one-half; average of 1, 0, and 1
is 2/3; and then 1/2 again. And it keeps on going like this. So we have
another sequence of numbers here. And actually when you look at that one, here
every second one is 1/2 and then the red ones here they also converge to 1/2; they have a limit of 1/2. So overall, the sequence that we're
generating here has a limit that is 1/2. So, in terms of this second sequence,
what we get is convergence. And it makes sense to say that, in some ways,
this series here does converge to 1/2 And there's actually the special name for this:
this sum is Cesaro convergent (if that sort of thing happens). If you've got a series,
you first do the sequence of partial sums and then you do the sequence of averages;
if the sequence of averages converges to a number, then this number
is called the Cesaro sum of the series that we're looking at. So the series is Cesaro
convergent. And actually this different sort of convergence, just different sort of
attaching a sum to a series makes a lot of sense and really, on a different
planet with different sort of civilizations, they may actually have
chosen this as their default definition of sum of a series. Could well be. This
sort of thing actually has just as nice properties as everything else. So, for
example, if a series converges in the standard way, then it will also be Cesaro
convergent. So that's really good. But obviously what we've seen here says there are
certain series that this guy here can sum that normal, you know, normal people can't, can't sum. Also, this sort of thing works. If you've got something that Cesaro sums
to A, then 5 times that is that one, in terms of Cesaro. And the adding business works, the sutracting
business works. And so what that means again is, you go back to the very
beginning If in your version of mathematics you're talking
about Cesaro sums, talking about something like this, then this calculation
here will actually give you the right answer. And it does it, right?
It gives us the answer One-half for (1-1+1...) and so on. Further in calculus, you actually learn about these things. They're actually very useful, so they not only make sense but also are very useful, even for the stuff
that, you know, we define as making sense. So, for example, before I told you that
when you've got something that has sum A, and there we've got another one with sum B, in the standard way... in the
standard way. Then the product not necessarily will converge. The product
sometime diverges. But when you add the product in the Cesaro way, you'll actually
always get what you expect: A times B So actually adding in the Cesaro way gets rid of
one of those not so nice things about the usual definition of sums. And there's other things
that Cesaro summation takes care of. There's something called Fejer's theorem, which is about Alright, so let's have a look at our insaneometer. Well, on planet Cesaro, if Ramanujan writes
the letter, he will actually only get an insanity score of 3, instead of an insanity
score of 4 that we had before. OK, now let's have a look at the next
sum here on our list, that was the 1-2 +3, and so on, sum. If you look at the partial sums, you get 1, -1, 2, and so on. So that doesn't work. Now, let's just do the Cesaro thing, so the averaging.
When we do the averaging, we get these guys here, and you see: well, that doesn't work, right,
that doesn't converge. So we've got every second one as 0; these guys here, the red ones,
they converge to, they have a limit of 1/2 OK, so if you actually step back here, step back
and look at the sequence of partial sums, you see every second one is a 0, and then
eventually the red ones will be indistinguishable from 1/2. OK, well,
there is 1/4 that we had before What do have 0 and 1/2 to do
with 1/4? Well, it's average, right, it's average, so we can actually kind of
repeat our game. So we're just do the averages of the averages now. So we do, you know, put down 1, and then we do the average of those two and then the average of those three, and then the average of those four, so this gives us another sequence of numbers, and that will
actually converge to 1/4. I'm not going to write the numbers now, but it's going to work.
And we can actually now choose these higher-order Cesaro thing to mean our sum.
Right, so on a different planet again you know, on a different planet, maybe on
the planet of the blue aliens, people define sums of, of series in terms of
this (first, second) third sequence of numbers that's associated with any series. Π‘ould do that. And a planet like this well, the insaneometer here will show just
a reading of 2, so on a plant like this, you know, that gets more and more
reasonable what, what Ramanujan does there. We can actually keep on going playing
this game; so we had like the first sequence (that's us), second sequence (that was my green alien
friend who prefers this sort of summation), then the blue alien friend who prefers the
next one down, but of course we can do more and more kind of averages; so the
next one would be the average of the average of the average, another sequence. And these methods
get more and more powerful, so you kind of go up and up and up and up and up, and you
can sum more and more of these series. But what about this guy here? Will we ever be able
to get to -1/12? Well, if you think about it, eh, no. Right? So we've got positive numbers here,
we do partial sums that's got to be positive numbers again. Average of positive numbers β again,
positive numbers; or whatever you do here, you'll never get anything that will get
into the minuses. Alright, for our super-sum, I have to tell you a little bit about the
functions. OK, so here we've got x^2 graphed over the positives. Now, we can
extend is nice smooth function into the negatives in infinitely many ways, and I've just
drawn like three of them. So there's, we can extend it as x^2, but also in many-many other ways, in a smooth way across 0 here. OK? Just keep that in mind. So far, we've been
talking about the real numbers. And the real numbers are usually represented by the real
number line. Now I also have to talk about the complex numbers, which is... usually
represented by the complex plane. Here there's some complex numbers. If you've
never heard of complex numbers, check out some of these videos here before you
watch the rest. Now, everything I've said about standard summing of series,
Cesaro sums, averages of averages sums Stays true if you're actually considering
series of complex numbers. You can also have functions... complex functions defined
on various bits of the complex plane. And some of them are actually super duper nice,
smooth; they're called analytic functions. OK, so if you've got, for example, a function
defined on the right part here of the complex plane (everything to the right of
the imaginary axis), how many different ways are there to extend this to the
left here, to the left part of the complex plane? Now, these analytic functions
are actually really-really-really nice, a lot nicer than real-valued functions. So, if
actually this continuation exists, then it's uniquely determined, so that's
something very-very special, and it's called the analytic continuation of
our analytic function. So these things are super duper nice. OK, let's have a look at
one of those things. So, here is an infinite series that depends on a variable Z, so
that Z can be any complex number. Right, now we can check for which complex
numbers does this thing converge, say, in a standard way. In the standard way
it converges everywhere here, in the yellow, so, for example, at 1, Z=1,
if we substitute, we get this series, and we've already seen this before in the Apu
paradox video. That sums to ln(2). Alright. Now, since you've watched this video,
you probably come up with an idea straight away: what if we don't sum in
the standard way but if we sum like the Cesaro way? And actually something
really-really nice happens β you get part of the analytic continuation defined like
this. So if you sum Cesaro, you get everything here to the right. It's really nice.
Among other things, you can figure out what's the value of the analytic
continuation here, at 0. And what you get? Well, at 0, we have... well, it's everywhere 1, 1, 1, 1, 1... so we get (1-1+...), and we know what
the Cesaro sum of that is β that's 1/2. OK, and then what about average
the averages? Well, that gets us everything up to here.
Let's just plug in (-1) here, so -1. Mm. Well, that sum, and so that tells us that
analytic continuation here is actually equal to 1/4. And then we do averages of averages of averages, and we get this guy there... And we keep on going like this. Actually, kind of nice
homework assignment: what do these question marks stand for? A bit hard :) Alright. So these things are actually useful, you know, for defining the analytic continuation of this
very important Dirichlet function. Now we'll change all the minuses to
pluses; that's the Riemann zeta function. So, the Holy Grail in mathematics β the Riemann
hypothesis β is all about this guy here. So it's very important to know what it is. Now,
the series (just with a standard summation) defines an analytic function here to
the right again of the purple line. So what does Cesaro give us? What does
averages of averages give us? What do all of the other things give us? Do they give us
the analytic continuation of this thing? Well, sadly, not really. You don't really
get anywhere with this. But there is an analytic continuation here, and let's just see what we get you formally
if we substitute (-1). So if we do (-1), we actually get our Ramanujan sum, super sum.
OK? OK, so what are we thinking now... Well, to start with, when we defined
these generalized sums, we took the first thing that came to mind here. But
what if there's different ways of generalizing the standard way of summing series?
Other ones, right, but more sophisticated ones. Maybe those will allow us to calculate
the analytic continuation of of this sum up there. And there actually are.
There are more complicated ones. For example, Ramanujan invented one. And then...
Well, since the are such sums, maybe it's possible to then calculate these sums using Ramanujan's that manipulation. And it's
actually true: the analytic continuation at (-1) is (-1/12). Now here is something very-very-very important. Those first couple of summations methods that I've talked about... They conceivably
could really be replacements of the standard way of summing things (on some
planet). Now, these more complicated summation methods, like Ramanujan's one (when
you look at it: it's got integrals, and it's got like sums in it, and special numbers), they
only work in special context. They work for special sorts of series. They're really-really
useful there, but they could never ever replace standard summation. If at any
point in time you get on your calculus test "what is 1+2+3+4 etc.", and you
answer "-1/12", almost certainly you'll get zero marks. Whenever you get something
like this, there's a standard way of interpreting a sum like this; and the
standard way involves, you know, the partial sums. Do the partial sums converge
or not over the reals β that's what's being asked here. Nothing else, OK? If you
give any other answer, it will be wrong. OK? Let's keep that in mind. So the last thing I want to talk about is again the geometric series because most of you
will be familiar with this, and most of you have been told to disregard anything
outside, you know, small values of Z. Now, this one actually works for complex
numbers too and defines a nice analytic function here in the unit circle. If you actually
now some using Cesaro, it will also give you things on the unit circle.
So it will actually give you the analytic continuation on the unit circle.
What else? Well, if you do averages of averages, actually doesn't
get you anymore. But there's an analytic continuation, of course,
of this function that's defined here. And this analytic continuation is actually
this sum here that we calculated before. So again, I mean, what comes out of there is
not nonsense; it defines something, it defines values of the analytic continuation
of the function that's defined by this guy here. But that's what it is.
Just started watching it, but he seems to be annoyed at all the horrible treatments this thing got on Numberphile and other places. He decided to do it right, and that's why the video is 30+ minutes long.
Should be interesting! Burkard doesn't disappoint and he does try to be rigorous but accessible.
I was really excited to see this posted because it's a topic that I've been wanting to understand ever since I'd heard about it on Numberphile. The video is very interesting and informative, but I feel like he glossed over what seems to be the most important part.
Everything he talked about built up to the Ramanujan summation, and then he just kind of says it's weird looking and is another form of analytic continuation. Well, all right, but that's the thing that actually allows us to get that -1/12 result, right?
I wish that there had been some attempt to break down that Ramanujan summation piece by piece so that the whole convergent/divergent > Cesaro summation > analytic continuation journey paid off more.
What are the individual parts of the Ramanujan summation? What does it actually tell us about the series it is applied to? Why is it useful? I've tried to read about it on wiki and elsewhere, but it's very dense and daunting when you don't have an extensive background in math.
So if I understood the video correctly it all boils down to how we decide to define convergence. In the standard way we are taught in calculus, many of these sequences are nonsensical. However, if we open up our ideas of convergence some of these do have meaning?
I have a couple of follow up questions? Are there other definitions of convergence, like Cesaro, not mentioned. Also, what are some of the practical applications of these alternative definitions of convergence?
Anyone have a pdf of the letter in the video?
I was a little skeptical going in to the video, as the other videos by Numberphile had left a lot of misunderstandings and drew a lot of misguided attacks. I thought this might be another awkward approach that could leave people still believing the manipulations were meaningless.
But he did a really good job. Really very solid.
He made it clear that the manipulations might be taken as the definition right off the bat. He pointed out that infinite sums were already meaningless using the definition of finite sums and how you already need a new definition. Love the aliens.
But then he started in about Cesaro summation and alternate summation theories. And then he worked in analytic continuation and the basis for why all this worked. And he showed how this actually preserves and extends the manipulation meaningfulness.
All nailing some of the crucial insights missing from that strange negative internet reaction to Numberphile's first video.
Personally, I would have liked a slightly different and extended ending, though. It was a wrong turn to start claiming Ramanujan summation was not one that could be considered a valid substitute like Cesaro and the aliens. What he needed to then discuss was Tauberian theorems and how summation theories are inclusively related. He needed to show how the manipulations project into the different theories, and their completeness properties. Because Ramanujan has a very natural place in relation to Cesaro and Abel and other theories. Once you start seeing series as places in analytic functions, you provide a natural semantics to the summation which has a well defined region of applicability.
All around, though, very happy to see this.
I criticized the video in my other comment, but I also would like to highlight this lovely fact from the video that I did not know: the product of two conditionally convergent series needn't be convergent, but it will always be CesΓ ro summable.
The studying limits of series requires one to look at the behavior of the sequence of partial sums. The standard definition for a convergent series is that the sequence of partial sums converges.
For example when you look at:
1/2 + 1/4 + 1/8 + 1/16 + 1/32 + ...
the sums get closer and closer to 1. In technical terms, the infinite sum is said to have a value of 1 because the sequence of the partial sums converges to 1.
That is the standard definition for a convergent series. Mathematicians are always looking for ways to tweak definitions to generalize results. One such generalization is Cesaro summation.
The Cesaro sum still relies on the behavior of the partial sums, but with a small hitch... it considers the limit of the average value of the partial sums rather than just the limit of the partial values themselves. At any given step you use all the partial sums up to that point and find the average...
Now this next part is really important... When using the Cesaro sum on a series which converges in the standard sense, that series still converges in the Cesaro sense and still converge to the same value...
Consider the sequence of partial sums from the above example:
1/2, 3/4, 7/8, 15/16, 31/32... (the partial sums are getting closer to 1 as you add more and more terms).
Now notice that if you start adding the partial sums together and computing the average of that sum, you are still getting closer and closer to 1.
(1/5)(1/2 + 3/4 + 7/8 + 15/16 + 31/32) = 129/160 (0.80625)
and if you go out 20 terms and consider the average of the first 20 partial sums the average will be about 0.95 and will be about 0.99 for the first 100 terms.
So, that is the Cesaro sum... and for convergent series (in the standard sense) it give the same value. BUT, the Cesaro sum allows us to assign values to additional series that do not converge in the standard sense..
For example the partial sums of Grandi's series 1-1+1-1+1-1+1-... alternate between 0 and 1, so the series does not converge in the standard sense. However, the limit of the average values of all those alternating partial sums does exist, and that limit is 1/2. So Grandi's series is said to converge to 1/2 in the Cesaro sense.
In a way, Cesaro summation smooths out noise in an otherwise convergent series to give a value. It is very useful for data that has some wiggle to it and is trying its darndest to converge... and it bears repeating, the Cesaro sum does not disturb any series that converge in the standard sense.
Now, I am gonna give an explanation by way of an analogy of how one can think 1+2+3... = -1/12. The analogy will contain much of the important mathematical intuition that is required to fully understand the curious sum of 1+2+3... = -1/12 but will not be the exact explanation. I will try to point out where the ideas of my explanation can be applied to the curious sum but as with all useful analogies it will fall short if examined too closely.
Consider the following real valued function:
f(x) = (x-3)/(x-3)
when you plug in numbers like x = 0, 1, 2 you get results like:
f(0) = -3/-3 = 1
f(1) = -2/-2 = 1
f(2) = -1/-1 = 1
and if you skip ahead and try numbers like x = 4, 5, 6 you get:
f(4) = 1/1 = 1
f(5) = 2/2 = 1
f(6) = 3/3 = 1
question... why did I skip ahead and not try x = 3? Well, when you try x = 3 you get:
f(x) = 0/0
which is undefined. So that is that, right? there is no way to think of f(x) = (x-3)/(x-3) so that 0/0 makes any sense, right?
Not so fast... if you were to graph f(x) it would be a nice graph everywhere except at x = 3 there would be an issue. What if we asked if there was a way to defined a value at x = 3 so that f(x) would have nice properties and still have all its old values... can we extend f(x) to some new f(x) that is nice?
Yes.. in fact, there is only one way to extend the definition of f(x) = (x-3)/(x-3) so that the extension is defined everywhere and continuous... and that is we assign the value f(3) = 1. We fill in the hole, so to speak.
Recap what happened, our original function f(x) was not defined for some values, but it turned out that f(x) was defined for enough values that there was a unique extension of f(x) to all real numbers and gave f(x) the nice property of being continuous. That extension requires that f(3) = 1.
Now back to the original problem with x = 3:
f(3) = (3-3)/(3-3) = 0/0
However, by extending f(x) to all real numbers (and keeping nice properties for f(x)) we are forced to use f(3) = 1. So now we know that:
f(3) = 0/0 = 1. So we 0/0 = 1, right?
No, 0/0 is nonsense from trying to put x=3 into our original f(x), but we can put x=3 into our unique extension of f(x) (the extension that filled the hole) and get f(3) = 1.
It turns out there is function g(z) called the Riemann zeta function defined on part of the complex numbers by the formula:
g(z) = 1-z + 2-z + 3-z + ...
Now g(-1) gives 1 + 2 + 3 +... which is nonsense, but it so happens that g(z) is defined on enough of the complex numbers so that there is a unique extension of g(z) to (almost) all complex numbers via analytic continuation and in that extension z = -1 gets a value that makes sense, in fact g(-1) = -1/12. So just as in f(3) = 0/0 = 1 we have g(-1) = 1 + 2 + 3 + ... = -1/12.
I hope that explains a little the seeming paradoxical nonsensical sum of 1+2+3... = -1/12 without having to directly appeal to Ramanujan Summation.
this has my seal of approval
r=2
f= a mess. Whatever the solution to 4f2 - 4f +- (f - 1)*sqrt(4f + 9) - 3 = 0 gives you