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)?
MY IDEAS
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.
MY GOAL
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.
Best Answer
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.