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$\displaystyle\sum_{n=1}^\infty \frac{1}{n^s}$ only converges to $\zeta(s)$ if $\text{Re}(s) > 1$.

Why should analytically continuing to $\zeta(-1)$ give the right answer?

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migrated from physics.stackexchange.com May 18 '11 at 4:34

This question came from our site for active researchers, academics and students of physics.

It's not clear (to me) what you mean by "the right answer". Since this was migrated from physics, I suspect that you have in mind the use of this summation in string theory. However, mathematically speaking, there is no "right answer" for summation techniques for divergent series; $-\frac{1}{12}$ is simply the answer that analytical continuation yields. See also en.wikipedia.org/wiki/…; . –  joriki May 18 '11 at 4:40
What does the phrase "right answer" mean? If you define zeta(s) in the usual way for s with real part greater than 1, it has an analytic continuation to s = -1 and the value there is -1/12. That does not mean the series defining zeta(s) for Re(s) > 1 converges at s = -1 with value -1/12 except in a hand-waving Euler kind of way. This question doesn't really have a mathematical meaning as far as I can tell. –  KCd May 18 '11 at 4:42
Someone should edit the Wikipedia section that says $1+2+3+... = \frac{-1}{12}$ en.wikipedia.org/wiki/Riemann_zeta_function#Specific_values –  zerosofthezeta Dec 5 '13 at 5:39
(For newcomers) there are a lot of linked questions now, but most relevant, IMO: i) on (problems with) finding infinite sums via 'shift & add', ii) on analytic continuation, iii) on assigning values to divergent series –  Grigory M Jan 12 '14 at 11:52

10 Answers 10

there are many ways to see that your result is the right one. What does the right one mean?

It means that whenever such a sum appears anywhere in physics - I explicitly emphasize that not just in string theory, also in experimentally doable measurements of the Casimir force (between parallel metals resulting from quantized standing electromagnetic waves in between) - and one knows that the result is finite, the only possible finite part of the result that may be consistent with other symmetries of the problem (and that is actually confirmed experimentally whenever it is possible) is equal to $-1/12$.

It's another widespread misconception (see all the incorrect comments right below your question) that the zeta-function regularization is the only way how to calculate the proper value. Let me show a completely different calculation - one that is a homework exercise in Joe Polchinski's "String Theory" textbook.

Exponential regulator method

Add an exponentially decreasing regulator to make the sum convergent - so that the sum becomes $$ S = \sum_{n=1}^{\infty} n e^{-\epsilon n} $$ Note that this is not equivalent to generalizing the sum to the zeta-function. In the zeta-function, the $n$ is the base that is exponentiated to the $s$th power. Here, the regulator has $n$ in the exponent. Obviously, the original sum of integers is obtained in the $\epsilon\to 0$ limit of the formula for $S$. In physics, $\epsilon$ would be viewed as a kind of "minimum distance" that can be resolved.

The sum above may be exactly evaluated and the result is (use Mathematica if you don't want to do it yourself, but you can do it yourself) $$ S = \frac{e^\epsilon}{(e^\epsilon-1)^2} $$ We will only need some Laurent expansion around $\epsilon = 0$. $$ S = \frac{1+\epsilon+\epsilon^2/2 + O(\epsilon^3)}{(\epsilon+\epsilon^2/2+\epsilon^3/6+O(\epsilon^4))^2} $$ We have $$ S = \frac{1}{\epsilon^2} \frac{1+\epsilon+\epsilon^2/2+O(\epsilon^3)}{(1+\epsilon/2+\epsilon^2/6+O(\epsilon^3))^2} $$ You see that the $1/\epsilon^2$ leading divergence survives and the next subleading term cancels. The resulting expansion may be calculated with this Mathematica command
1/epsilon^2 * Series[epsilon^2 Sum[n Exp[-n epsilon], {n, 1, Infinity}], {epsilon, 0, 5}]

and the result is $$ \frac{1}{\epsilon^2} - \frac{1}{12} + \frac{\epsilon^2}{240} + O(\epsilon^4) $$ In the $\epsilon\to 0$ limit we were interested in, the $\epsilon^2/240$ term as well as the smaller ones go to zero and may be erased. The leading divergence $1/\epsilon^2$ may be and must be canceled by a local counterterm - a vacuum energy term. This is true for the Casimir effect in electromagnetism (in this case, the cancelled pole may be interpreted as the sum of the zero-point energies in the case that no metals were bounding the region), zero-point energies in string theory, and everywhere else. The cancellation of the leading divergence is needed for physics to be finite - but one may guarantee that the counterterm won't affect the finite term, $-1/12$, which is the correct result of the sum.

In physics applications, $\epsilon$ would be dimensionful and its different powers are sharply separated and may be treated individually. That's why the local counterterms may eliminate the leading divergence but don't affect the finite part. That's also why you couldn't have used a more complex regulator, like $\exp(-(\epsilon+\epsilon^2)n)$.

There are many other, apparently inequivalent ways to compute the right value of the sum. It is not just the zeta function.

Euler's method

Let me present one more, slightly less modern, method that was used by Leonhard Euler to calculate that the sum of integers is $-1/12$. It's of course a bit more heuristic but his heuristic approach showed that he had a good intuition and the derivation could be turned into a modern physics derivation, too.

We will work with two sums, $$ S = 1+2+3+4+5+\dots, \quad T = 1-2+3-4+5-\dots $$ Extrapolating the geometric and similar sums to the divergent (and, in this case, marginally divergent) domain of values of $x$, the expression $T$ may be summed according to the Taylor expansion $$ \frac{1}{(1+x)^2} = 1 - 2x + 3x^2 -4x^3 + \dots $$ Substitute $x=1$ to see that $T=+1/4$. The value of $S$ is easily calculated now: $$ T = (1+2+3+\dots) - 2\times (2+4+6+\dots) = (1+2+3+\dots) (1 - 4) = -3S$$ so $S=-T/3=-1/12$.

A zeta-function calculation

A somewhat unusual calculation of $\zeta(-1)=-1/12$ of mine may be found here:


The comments are in Czech but the equations represent bulk of the language that really matters, so the Czech comments shouldn't be a problem. A new argument (subscript) $s$ is added to the zeta function. The new function is the old zeta function for $s=0$ and for $s=1$, it only differs by one. We Taylor expand around $s=0$ to get to $s=1$ and we find out that only a finite number of terms survives if the main argument $x$ is a non-positive integer. The resulting recursive relations for the zeta function allow us to compute the values of the zeta-function at integers smaller than $1$, and prove that the function vanishes at negative even values of $x$.

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The fact that other methods also yield the result $-\frac{1}{12}$ doesn't make any of the comments under the question "incorrect". You interpreted "right answer" to mean the one that works in physical applications; that's fine (and I suggested in my comment that it was probably meant that way), but that doesn't make it the right answer in any mathematical sense of the term, and nor does the consistency of several summation methods. I'm not aware of any theory of summation of divergent series that provides a definition of what "the right answer" for a resummation is. –  joriki May 18 '11 at 6:04
@Luboš: It seems we have a difference of opinion about the use of "right" in this context. I don't object to you using "right" in the sense that you're using it; what I object to is labelling other views as "incorrect". To call something "incorrect", you'd need a definition of what makes an answer the right answer. As long as you don't have one, it's a matter of taste that you call the consistent result of several different methods "the right answer" and I don't. –  joriki May 18 '11 at 6:47
@Luboš: I object to your combative tone. None of the comments under the question "lead to incorrect conclusions in all of physics". Noone here has questioned the fact that $-1/12$ is the answer that works in physics. There is merely a difference of opinion about whether it should be called "the right answer". I accept that you call it that, and I'd appreciate if you would accept that I don't, instead of labelling that as "shallow". –  joriki May 18 '11 at 7:26
@Luboš: I too object to your label of comments by others as "shallow". Look at the question: it asks specifically about analytically continuing and writes the value as $\zeta(-1)$. That is what I was responding to in my comment. I am well aware that there are other methods of "deriving" the value -1/12 and I had sketched Euler's technique a while ago on Mathoverflow (see my 2nd answer to mathoverflow.net/questions/13130/…). You answered a better question than the one that was asked. It's not a reason to say more focused answers are wrong. –  KCd May 18 '11 at 7:59
@Luboš: OK, if that's how you see it, I don't think there's any point in exchanging further arguments. I would just suggest that you look at the upvotes for the comments both here and under the question. Truth isn't democratic, but the numbers might motivate you to consider the possibility that there is more than one valid viewpoint in this case. –  joriki May 18 '11 at 10:09

Here is a variant on Lubos Motl's answer:

Let $S = \sum_{n=1}^{\infty} n$. Then $S - 4 S = \sum_{n = 1}^{\infty} (-1)^{n-1} n.$ We will evaluate this latter expression with a regularization similar to Lubos Motl's.

Namely, consider $$\sum_{n=1}^{\infty} (-1)^{n-1} n t^n = -t \dfrac{d}{dt} \sum_{n=1}^{\infty} (-t)^n = -t \dfrac{d}{dt} \dfrac{1}{1+t} = \dfrac{t}{(1+t)^2}.$$
Letting $t \to 1,$ we find that $-3 S = \dfrac{1}{4}$, and hence that $S = \dfrac{-1}{12}.$

To see the relationship between this approach and Lubos Motl's, note that if we write $t = e^{\epsilon},$ then $t\dfrac{d}{dt} = \dfrac{d}{d\epsilon},$ so in fact the arguments are essentially the same, except that Lubos doesn't perform the initial step of replacing $S$ by $S - 4S$, which means that he has the pole $\dfrac{1}{\epsilon^2}$ which he then subtracts away.

As far as I know, this trick of replacing $\zeta(s)$ by $(1-2^{-s+1})^{-1}\zeta(s)$ is due to Euler, and it is a now standard method for replacing $\zeta(s)$ by a function which carries the same information, but does not have a pole at $s = 1$. The evaluation of $\zeta(s)$ at negative integers by passing to $(1-2^{-s+1})\zeta(s)$ and then performing Abelian regularization as above is also due to Euler, I believe. It is easy to see the Bernoulli numbers appearing in this way, for example.

Of course, taken literally, the series $\sum_{n=1}^{\infty} n$ diverges to $+\infty$, so any attempt to assign it a finite value will involve some form of regularization. Analytic continuation of the $\zeta$-function is one form of regularization, and the Abelian regularization that Lubos Motl and I are making is another. I can't quote a precise theorem to this effect (although maybe others can), but with such a simple expression as $\sum_{n = 1}^{\infty} n,$ I'm reasonably confident that any sensible regularization will necessarily yield the same value of $-1/12$. (Lubos Motl makes the same assertion in his answer.)

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@Matt: My first example was flawed, but I think my general point still stands that a satisfactory theory of summation of divergent series would have to explain not only why several seemingly different methods yield the same result, but also why other methods don't. Here's a new example (hopefully less flawed): $S-2S=1+3+5+\dotso$ Now we can subtract $2S$ again, and depending on whether we start substracting the $2$ from the $3$ or the $1$, we get $1+1+\dotso$ or $-1-1-\dotso$ First, these should be different. Also, $1+1+\dotso=\zeta(0)=-1/2$, so that would yield $S-4S=-1/2$, $S=1/6$. –  joriki May 18 '11 at 6:57
@Matt: (I'm not saying that this can't be explained, or that the results that do coincide do so by chance; to the contrary, I find it intriguing how many of them coincide; all I'm saying is that the ones that don't coincide are also part of the picture, and a satisfactory theory (one that would in my opinion allow us to speak of "the right answer") would have to have something to say about which of these methods are "valid", which aren't and why.) –  joriki May 18 '11 at 7:31
@Luboš: Instead of insisting on using terms like "simple errors" "in reality" and "wrong interpretations", it would be more helpful if you could specifically point out in which sense you believe that your calculation of $S-2S-2S$ has more merit than mine. As I emphasized, it seems quite likely to me that there is such a difference; but I was looking for an explanation. How did I "randomly clump" terms in my calculation? In which framework does your combination of the terms appear more systematic than mine? (Not rhetorical questions.) –  joriki May 18 '11 at 9:55
@joriki: Dear Joriki, I deleted my comment that was made in reference to your deleted comment. As for your revised example (obtaining $\zeta(0)$), there is a "graded" aspect to $S$, which is being disturbed in your computation. (This is what Lubos is referring to when he writes that you "randomly clump" terms.) In terms of Lubos's answer, this disturbing of the grading is not allowed because of dimensional considerations. Mathematically, all I can think to say right now is that both $\zeta$-regularization and Euler regularization involve the graded aspect of $S$, and preserve it. Regards, –  Matt E May 18 '11 at 11:58
I don't like the first step of your derivation very much, where you compute $S-4S$. Implicitly, this uses the fact that the sums $1+2+3+\cdots$ and $0+1+0+2+0+3+0+\cdots$ have the same value. While true, this is not immediately obvious. For instance, the sum $1+0+2+0+3+0+\cdots$ has a different value (namely $1/24$). –  Axel Boldt Mar 9 '14 at 3:10

What a great method! I tried a few... $$ \sum_{n = 1}^{\infty} \left(\frac{1}{(2 n)^{s}} - \frac{1}{(2 n - 1)^{s}}\right) = \zeta (s) \Bigl(2^{(1 - s)} - 1\Bigr) $$ put $s=-1$ to get $\sum_{n = 1}^{\infty} 1 = -\frac{1}{4}$

or $$ \sum_{n = 1}^{\infty} \left(\frac{1}{(2 n + 1)^{s}} - \frac{1}{(2 n)^{s}}\right) = \zeta (s) - \frac{2 \zeta (s)}{2^{s}} - 1 $$ put $s=-1$ to get $\sum_{n = 1}^{\infty} 1 = -\frac{3}{4}$

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You should read Terence Tao's article on this "oddities". –  Pedro Tamaroff Jun 11 '12 at 17:37
For clarity: I'm assuming you mean terrytao.wordpress.com/2010/04/10/… (which I found through JavaMan's comment on the question. –  Marnix Klooster Jan 16 '14 at 20:04

Here is a direct computation of $\zeta(-1)$ taken from this answer:

Multiply equation $(1)$ from this answer by $x+1$, then integrate by parts twice, to get $$ \begin{align} (1-2^{1-x})\zeta(x)\Gamma(x+2) &=\int_0^\infty\frac{(x+1)xt^{x-1}}{e^t+1}\mathrm{d}t\\ &=\int_0^\infty\frac{(x+1)t^xe^t}{(e^t+1)^2}\mathrm{d}t\\ &=\int_0^\infty\frac{t^{x+1}(e^{2t}-e^t)}{(e^t+1)^3}\mathrm{d}t\tag{1} \end{align} $$ Now we can plug in $x=-1$ into $(1)$ to get $$ \begin{align} (1-2^2)\zeta(-1)\Gamma(1) &=\int_0^\infty\frac{e^{2t}-e^t}{(e^t+1)^3}\mathrm{d}t\\ &=\int_1^\infty\frac{u-1}{(u+1)^3}\mathrm{d}u\\ &=\int_1^\infty\left(\frac1{(u+1)^2}-\frac2{(u+1)^3}\right)\mathrm{d}u\\ &=\frac14\tag{2} \end{align} $$ Since $(1-2^2)\Gamma(1)=-3$, $(2)$ says that $$ \zeta(-1)=-\frac1{12}\tag{3} $$ Naively plugging $n=-1$ into $$ \zeta(n)=\frac1{1^n}+\frac1{2^n}+\frac1{3^n}+\dots\tag{4} $$ yields $$ -\frac1{12}=1+2+3+4+\dots\tag{5} $$ However, the series on the right of $(5)$ is obviously divergent by the Term Test.

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A good explanation of how this is true and why and where it is useful and significant is beautifully summarized in 24 by John Baez (as part of the 19/09/08 Rankin Lectures) and yes, I do agree that this is wierd.

The following is the same Euler's method described by Luboš Motl and Matt E in the other answers.

Let us consider the infinite geometric progression " $1, x, x^2, x^3, x^4, \dots$ "

The sum of this progression can be found by the formula $S_\infty = \large \frac{a}{1-r}$,

$$\implies 1 + x + x^2 + x^3 + x^4 + \dots = \frac{1}{1-x}$$ Differentiating with respect to $x$, $$ \implies 0 + 1 + 2x + 3x^2 + 4x^3 + \dots = \frac{1}{(x-1)^2}$$ Let $x = -1$, (Note: In the LHS, you get the alternating counterpart of the natural numbers series) $$\implies 1-2+3-4+5-6+...= \frac{1}{4} $$

$$ \bbox[5px,border:2px solid green]{\therefore \sum\limits_{i=1}^\infty n(-1)^{n-1} = \frac{1}{4}} $$

Subtracting this from $\sum\limits_{i=1}^\infty n$, $$\sum\limits_{i=1}^\infty n -\sum\limits_{i=1}^\infty n(-1)^{n-1} =4+8+12+...=4\sum\limits_{i=1}^\infty n$$ $$\implies 3\sum\limits_{i=1}^\infty n =-\sum\limits_{i=1}^\infty n(-1)^{n-1}=-1/4$$ $$\bbox[5px,border:2px solid lime]{\therefore \sum\limits_{i=1}^\infty n=-\frac{1}{12}}$$

This intuitive reasoning was made famous in a video by the physicist Phil Plait of Numberphile. It is heavily criticized for being sloppy but it shows a simple method that has fooled a large population of its high school audience (as noted in the comments).
Funfact: Googling -1/12 doesn't give you any results

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That is not a proof, since you are manipulating expressions to which no meaning is attached. One can attach meaning to (some of) them, but that is a somewhat technical thing, and quite non-obvious. (It may be the case that you do know how to cope with these technical difficulties, and that when you write, say, «$1+2+3+4+\cdots$» you h ave in mind something concrete and meaningful, but in that case you should make it explicit) –  Mariano Suárez-Alvarez Jan 18 '14 at 2:37
This proof was featured today in a video posted to Reddit. –  Steven Gubkin Jan 18 '14 at 3:31
Maybe not a proof but still a nice and simple way how to fool a random high-school audience. Thumbs up. –  Jeyekomon Jan 18 '14 at 17:05
@BalarkaSen, if you take the trouble of looking at history of edits in this post, you will notice that at the time I made the comment above the answer was of a somewhat different nature... –  Mariano Suárez-Alvarez Jan 19 '14 at 6:27
You manipulate that infinite sum as if it were finite, like when you assume that $\sum_{n=1}^\infty n-4\sum_{n=1}^\infty n=-3\sum_{n=1}^\infty n.$ But that's impossible, since $\infty-\infty$ is undefined. –  Hakim May 16 '14 at 20:52

The notation "$1+2+3+\cdots$" is as meaningless as "$1/0$". If you treat such notation as though it defined a real number and conformed in its syntax to the rules of formation for genuine real numbers, you can easily "prove" it to equal any number you like, including $-1/12$.

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Dear John, Your claim that this notation is "meanlingless" seems unnecessarily absolute, in light of the answers explaining that it does in fact admit meaningful interpretations. Regards, –  Matt E Sep 8 '12 at 19:11
Dear Matt: Sure, it admits meaningful interpretations, for some people. But they are not consistent. –  John Bentin Sep 9 '12 at 12:03
The fact that so many different methods which connect to deeper waters in mathematics all lead to the same sum suggest that, even though there is currently no way to make it precise, there might be a rigorous consistent theory of divergent sums which no one has made fully precise yet. To me, these are some of the most exciting things in mathematics: having only a glimpse of something great just beyond the horizon. –  Steven Gubkin Jan 18 '14 at 3:29
@StevenGubkin Replace "there might be a rigorous consistent theory of divergent sums which no one has made fully precise yet" by "there exists a rigorous consistent theory of divergent sums, made fully precise more than 80 years ago". As Littlewood explains in the preface of Hardy's treatise, "[I]n the early years of the century the subject [Divergent Series], while in no way mystical or unrigorous, was regarded as sensational, and about the present title, now colourless, there hung an aroma of paradox and audacity"... only these were the first years of the 20th century, not the 21st. –  Did Jan 18 '14 at 21:50
I think everything said in this post is true... but none of it explains why $-1/12$ is a better choice of "sum" than any other real number. (If you think $-1/12$ is no better than any other real number here, well, just read the other answers.) –  Jesse Madnick Jan 22 '14 at 1:36

If the following were true: $$\sum_{n=1}^\infty{n}=-\frac1{12}\tag{hypothesis}$$ then we would expect the following: $$\lim_{n\to\infty}\frac{n(n+1)}2=-\frac1{12}\tag{expectation}$$ which is the formula for the infinite triangular number limit. Unfortunately this is a result that we do not get when the limit is correctly taken. The correct value is $$\lim_{n\to\infty}\frac{n(n+1)}2=\frac{\infty(\infty+1)}2=\frac{\infty^2+\infty}2=\frac\infty2=\infty\neq-\frac1{12}$$ This sort of mathematical sleight of hand, smoke and mirrors, pulling a finite negative rabbit out of an empty positively infinite hat does not impress me; worse yet, it gives legitimate, observable, repeatable mathematics a bad name.

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Great answer! You should keep the "lim n to infinity" operator in front of every expression until the end instead of simply plugging in infinity... –  zerosofthezeta Mar 13 '14 at 3:33
Not sure that is expectable. I made a question related to it. –  JMCF125 Apr 27 '14 at 20:10
I can't tell if this was meant to be a joke, or not. $∞²$ is a nice touch. –  primo May 29 '14 at 10:24
Sum of all natural numbers=negative fraction.This mathematical truth is too bitter. Isn't it violating the basic rules of mathematics(i.e. sum of positive natural numbers is positive). This mathematical phenomenon or behaviour is weird and incomprehensible. Doesn't this kind of situation tell us that human mind is insufficient to understand everything; may be thought is not capable enough to grasp the nature of every truth. –  Sara Tancredi Jun 5 '14 at 23:47
@GottfriedHelms, did you just divide by 0 ? –  GinKin Jun 17 '14 at 19:44

A nice alternative representation, by a double sum.

Consider the infinite square array where the rowsums and the (formal) column-sums are also given in closed forms $$ \small \begin{array} {r|rrrrr|rr} & & & & & & \text{rowsums} \\ \hline & 1/0! & \log(1)/1! & \log(1)^2/2! &\log(1)^3/3! & \cdots & = e^{\log(1)}&=1 \\ & 1/0! & \log(2)/1! & \log(2)^2/2! &\log(2)^3/3! & \cdots & = e^{\log(2)}&=2 \\ & 1/0! & \log(3)/1! & \log(3)^2/2! &\log(3)^3/3! & \cdots & = e^{\log(3)}&=3 \\ & \vdots & & & & \ddots & \vdots & \vdots\\ \hline \text{colsums:} & \zeta(0) & -\zeta(0)'/1! & \zeta(0)''/2!& -\zeta(0)^{(3)}/3! & \cdots &&=\zeta(-1) \end{array} $$ and the row-sum of the derivatives of the $\zeta()$ at $0$ in the bottom row can numerically be written as $$ \small -0.5 +0.9189... -1.003178...+1.000785...-0.999879...+1.0000019... \pm \cdots $$ If we split the terms and do two series we find $$ \Tiny \begin{array} {} -1 &+1 &-1 &+1 &-1 &+1 &\pm &\cdots &=-1/2\\ +0.5 &-0.0810... &-0.003178...&+0.000785...&+0.000120...&+0.0000019... &\pm &\cdots &= 5/12\\ \end{array}$$ Here we must now resort to the earlier definition of the divergent sum of the alternating units (in the first row), but the second row is convergent and can conventionally be summed. The sum of the two rowsums $\small -1/2 + 5/12 = -1/12$ gives the expected result.

However, that fiddling with double-sums, when non-convergent-series are involved (as are the columnsums in the above matrix and the splitting in the numerical expression) must explicitely be justified as "legal"/algebraically consistent operation.
But because I encounter it not frequently, that such non-convergent double-sum schemes come out with the expected result without further ado, I think this is a specific nice observation here.

And the more algebraically manipulations come out to be consistent with the assumed value of a divergent series, the more is the hypothese acceptable, that this should be taken as the canonical numerical replacement for the series-expression ( like we do it with the rational fraction for the geometric series even in the divergent case (except for that with quotient $q=1$)) .

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This infinite series is ultimately divergent because $$ 1+2+3+4+\cdots=\sum\limits_{k=1}^{\infty} k = \lim\limits_{n\to\infty} \sum\limits_{k=1}^{n} k = \lim\limits_{n\to\infty} \frac{n(n+1)}{2} = \infty $$ The value of $-\frac{1}{12}$ is assigned to the infinite series by the methods of zeta function regularization and several others. If you'd like to understand why someone would assign a value to a divergent series, then have a look at this question.

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