A way to evaluate $\sum_{n=0}^\infty(-1)^{n+1}n$

Question

The binomial expansion of $(a+b)^{-2}$ is given as

$$(a+b)^{-2}=\sum_{n=1}^\infty(-1)^{n+1}na^{-1-n}b^{n-1}\tag{I think}$$

And when $a=b=1$,

$$2^{-2}=\sum_{n=1}^\infty(-1)^{n+1}n=1-2+3-4+\dots$$

So I was wondering if this were a way to evaluate the divergent summation in a ramanujan sort of meaning.

Answer 1

Yes this is Abel Summation (Regularization) because it comes from defining:

$$f(z)=\sum_{n=0}^\infty a_nz^n$$ and evaluating:

$$A:=\lim_{z\rightarrow 1^{-}}\sum_{n=0}^\infty a_nz^{n}.$$

In your case:

$$f(z)=\frac{1}{(1+z)^2}=\sum_{n=1}^\infty ((-1)^nn)z^n.$$

You can find other ways to regularize your sum here.