Solving $\sum_{n=1}^{\infty}(-1)^{n+1}(\frac{1}{n})$ using Abel's Theorem

Question

I am looking for some help with an real analysis problem that I have.

Problem: Find the sum of the series $\sum_{n=1}^{\infty}(-1)^{n+1}(\frac{1}{n})=1-(1/2)+(1/3)-(1/4)+\ldots $

What I have so far: My intuition suggests that I could use Abel's theorem which states that if $G(x)=\sum_{k=0}^{\infty}a_{k}x^{k}$ is a power series with real coefficients converges and the radius of convergence is 1, then $\lim_{x\to1^{-}}G(x)=\sum_{k=0}^{\infty}a_k$.

So then I tried to rewrite the power series given in the problem to have a starting index of 0. $\sum_{n=1}^{\infty}(-1)^{n+1}(\frac{1}{n})=\sum_{k=0}^{\infty}(-1)^{k}\frac{1}{k+1}=\lim_{x\to1^{-}}\ln(1+x)=ln(2)$ which I think would hold because the radius of convergence of $\ln(1+x)$ is 1. I am not sure if this is the right idea or not.

Answer 1

Note that $$\begin{split} \sum_{n=1}^{\infty}(-1)^{n+1}\frac{1}{n} & =\sum_{n=0}^{\infty}(-1)^{n}\frac{1}{n+1}\\ &=\sum_{n=0}^{+\infty}\int_0^1(-1)^{n}x^ndx\\ &=\int_0^1 \sum_{n=0}^{+\infty}(-x^n)dx \end{split} $$ where the integral and sum can be swapped by the dominated convergence theorem. Indeed $$\left|\sum_{n=0}^N(-x)^n \right|=\frac{1-(-x)^{N+1}}{1+x}\leq\frac{1}{1+x}$$ Consequently $$\sum_{n=1}^{\infty}(-1)^{n+1}\frac{1}{n}=\int_0^1\frac 1 {1+x}dx = \ln 2$$

Edit: I realized that the OP was explicitly asking for using Abel's theorem. So I'm not exactly answering the question. Sorry about that. Leaving this here, in case it's useful.

Answer 2

I would think you could use the Leibniz test (a special case of Abel's theorem), since the series is alternating and decreasing.

That is, the series alternates and $\mid a_{n+1}\mid\lt\mid a_n\mid$. So it converges.

Your idea appears to be correct. See Abel's theorem .