This is a strictly preliminary question. I hope to elicit some discussion/s which will lead to a more acceptable form for the question on this site.

I'm trying to understand how the study of the following infinite series:

$1 + x^2 + x^4 + x^6 + \cdots$ [1]

and complex numbers in general are related.

Specifically, given the relation:

$1 + x^2 + x^4 + x^6 + \cdots$ = $(1 - x^2)^{-1}$ [2],


$1 + 2^2 + 2^4 + 2^6 + \cdots$ = -$\frac{1}{3}$ [3]

in what way do complex number concepts come into play when trying to understand the 'intricacies' of equation [3] above?

(I've come to gain some understanding of the equations above; namely:

(1) the relation [2] is only applicable when |x| < 1;

(2) Euler was generally happy to use relations as [3] in his studies.

But, specifically, in trying to study this problem, where and how are complex numbers employed to come to a deeper understanding of the question?)

Request to potential answerers: I have a very minimal background in real analysis, and almost none of complex analysis. If you may be kind enough to answer / respond so I may understand your answer / response that would be greatly appreciated. :)

  • 4
    $\begingroup$ This falls under the purview of series regularization and summability theory. Look at e.g. Abel, Cesàro or Ramanujan summation. The connection to complex analysis comes through analytic continuation of the sort of functional expansions defining any particular summation method. $\endgroup$
    – anon
    Sep 30, 2011 at 11:23

2 Answers 2


The connection between $1 + x^2 + x^4 + x^6 + \cdots=(1 - x^2)^{-1}$ for $|x|<1$ and $1 + 2^2 + 2^4 + 2^6 + \cdots=-\frac{1}{3}$, which is the extention of $(1 - x^2)^{-1}$ to $x=2$ (where the series doesn't converge), is called analytic continuation.

  • $\begingroup$ Can you provide a summary which is comprehensible to a 2nd-3rd year undergrad with modest knowledge of real analysis, and almost none of complex analysis (apart from complex numbers and some basic identities)? $\endgroup$
    – UGPhysics
    Sep 30, 2011 at 11:37
  • 3
    $\begingroup$ @UGPhysics: Remember the time when the exponents in $x^n$ were defined only for $n$ a positive integer? Then we tried to find out what happens if $n$ is $0$ or negative. Then we tried to find out what happens when $n$ is some fraction. Much later in calculus, we made sense of what $x^n$ means for real (or even complex) $n$. That's analytic continuation there. You try to find an expression that makes sense outside the original domain of your function, but still agrees with your function in the original domain. $\endgroup$ Sep 30, 2011 at 12:16

The core idea, I believe, is that analytic functions are more general objects than power series. Sure, you can represent an analytic function locally by a convergent power series, but it is not guaranteed to converge outside of a certain scope (namely, an open disk or abscissa), even though the function exists elsewhere too. It's like how you can see a very small area around a character in a video game but otherwise can't see the entire world all at once (although world maps on the start menu put a damper on that analogy).

However, the formal characterization of the terms in a series (for example, seeing $2^{-n}$ and considering it a special case of $x^n$ at $x=1/2$) is enough to encode all of the local behavior of a function, and amazingly enough if you have all of the local information possible about a complex-analytic function at a certain point (namely, all of its derivatives there), then this information determines the function's behavior everywhere - but only "determined" in the sense of fixing the function's values everywhere, ensuring no extra degrees of freedom for outside behavior, not in the sense of explicit construction. A power series is not always up to the task of representing that same behavior everywhere.

Take the geometric series $\frac{1}{1-x}=1+x+x^2+x^3+\cdots$ for example. If you plot $1/(1-x)$, you get

$\hskip 1.7in $ geo

which shows $y=1/2$ at $x=-1$ even though $1-1+1-1+1-\cdots$ does not converge. IOW, $$\lim_{x\to-1^+}\left(\sum_{n=0}^\infty x^n\right) \quad\ne\quad \sum_{n=0}^\infty \left(\lim_{x\to-1^+}x^n\right).$$

So the power series representation is not capable of going to or any left of $x=-1$, even though there is a more general, and unique (up to complex-analyticity and monodromy), object (the analytic function $1/(1-x)$) that agrees with the power series on the interval $(-1,1)$ and also extends beyond. This is why we say the function analytically continues the original expression of the geometric series. You can also look at series representations of $1/(1-x)$ elsewhere, for example the Taylor expansion about $x=-1$ gives

$$\frac{1}{1-x}=\frac{1}{2}+\frac{x+1}{4}+\frac{(x+1)^2}{8}+\cdots,$$ valid strictly over the interval $(-3,1)$. And then you could give another power series to the left of that, and then another to the left of that, and so on. Thus analytic continuation can extend the geometric series indefinitely to the left. But of course, both the geometric series and the actual function $1/(1-x)$ have a pole at $x=1$, so looking strictly at the real line we can't go any further right. This, however, does not stop us when we look at the complex plane at large.

When in the complex plane, we find that a power series originally defined on the reals that converged only on an interval now converges only on an open disk. However, not every point on the outside of such a disk will necessarily be a pole of the general function that the power series converges to, so we can take an escape path out of narrowly defined defined region just like we extended the geometric series to the left of $-1$, and we end up broadening our general function more and more if we extend indefinitely. Wikipedia has an illustration of this:

$\hskip 0.6in $ logarithm

You will notice a slight issue in the above: if you go through the process of analytic continuation around a pole (here $z=0$), you might not end up at the same complex number that the original series evaluated to. This means that the general function must be defined with a branch cut (or else be studied by more sophisticated theory).

Now back to divergent series. If you have one and you want to use an analytic function to regularize it, you have to make a choice of what general form you want it to be an instance of. For example, with $1+2+4+8+\cdots$, you could choose to look at the geometric series formally evaluated at $x=2$, or you could say it is $e^{-0s}+e^{-1s}+e^{-2s}+e^{-3s}+\cdots$ evaluated at $s=-\log2$. It is not generally guaranteed that you will get a unique answer - it may depend on what your choice of general function is - but there are standard choices that can be invoked. One method is zeta regularization, where in order to evaluate the infinite sum $a_1+a_2+a_3+\cdots$ we define a zeta function $$\zeta_A(s)=\frac{1}{a_1^s}+\frac{1}{a_2^s}+\frac{1}{a_3^s}+\cdots$$ and then analytically continue it out of its abscissa of convergence and evaluate it at $s=-1$. (This is why $1+2+3+\cdots=\zeta(-1)=-1/12$ in an Euleresque way.) It also turns out that in addition to taking a series and turning it into an instance of a more general function's divergent power series from another area of convergence, one can perform operations on the partial sums in order to evaluate to a finite expression - these forms of regularization are called summability methods. (I'll try and expand on these later...)

  • $\begingroup$ Hi @anon, your answer seems well-thought-out and valuable to someone with complex analysis under their belt, but I'm afraid currently it's somewhat over my head. (I'm still giving it +1 since I will review it once I gain sufficient knowledge / background in the subject.) - cheers $\endgroup$
    – UGPhysics
    Oct 5, 2011 at 12:33

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