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Suppose we are given a series that diverges. That's right, diverges. We may interest ourselves in the limiting function(s) of its behavior.

For instance, given the power series:$$\frac{1}{1+x} = 1 - x + x^2 - x^3 + \dots$$

I am interested in finding the sum of the coefficients for $x^n$ and $x^{n+1}$ as $n$ approaches infinity. This should be fairly obvious as to what it is, but is there some way that we could do this for a general alternating series that essentially converges to two values, or more importantly, for 3 or more values (i.e. not exactly an alternating series, but one that cycles through a set of values at infinity)?


I guess that this can somehow be accomplished similar to finding the limiting behavior for a convergent series. I thought that I knew how to do this, but I forgot what I thought was an easy way.


I really would like to know if it's possible to apply a function to the values at the limit. If it is, of course, I'd like to know how to do this. This may allow us to determine what the values actually are by using multiple functions.

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The limit of the means of the partial sums (if x=1)? For bigger x you can do Euler- or Borel-summation. There is good introductory material at wikipedia which covers your question. – Gottfried Helms Feb 16 '12 at 1:01
up vote 1 down vote accepted

Note that the usual definition of the infinite sum is a very different kind of thing from an ordinary finite sum. It introduces ideas from topology, analysis or metric spaces which aren't present in the original definition of sum. So when generalising from finite sums to infinite sums we have quite a bit of choice in how these concepts are introduced and it's something of a prejudice to call the standard method from analysis the sum.

There are quite a few different approaches to summing infinite series, many of which give finite values in places where the usual summation method fails. Examples are Abel summation, Borel summation and Cesàro summation. The mathematician Hardy wrote an entire book on techniques to sum divergent series.

These alternative summation methods aren't just a mathematical curiosity. They play a role in physics where they can help sum some of the series that arise from sums over Feynman diagrams. These results often agree with results computed by more orthodox methods.

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Thanks for that link. Do you mind giving yourself a proper name or nickname? – Pedro Tamaroff Feb 16 '12 at 1:52

One common method for finding a "sum" of a divergent series is to multiply each term by $x^n$ and to let $x\to1$. For example, the alternating series $$ 1-2+3-4+5-6+\dots\tag{1} $$ turns into $$ 1-2x+3x^2-4x^3+5x^4-6x^5+\dots=\frac{1}{(1+x)^2}\tag{2} $$ so that $(1)=\frac14$.

Another trick can be used to get the "sum" of $$ 1+2+3+4+5+6+\dots\tag{3} $$ Note that subtracting $4\times\!(3)$ from $(3)$ yields $$ \begin{align} &\hphantom{=\;}1+2+3+4+5+6+\dots\\ &\hphantom{=\;}\hphantom{1}-4\hphantom{+3}\,\ -8\hphantom{+5}\,-12-\dots\\ &=1-2+3-4+5-6\tag{4} \end{align} $$ Thus, $-3\times\!(3)=(4)=\frac14$. Therefore, $(3)=-\frac{1}{12}$; this also happens to be $\zeta(-1)$.

These are just a couple of the ways to assign a "value" to a divergent series.

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Add to rob's answer. Note that for your function, you have

$$\frac{1}{2} = 1-1+1-1+1-1+\cdots$$

You may find this helpful too.

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That paper talks about the Euler Series Transformation, which is quite useful in many situations. I have written up a rigorous proof, but it was written up for sci.math, so it is in ASCII. – robjohn Feb 16 '12 at 13:44

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