Reading my old analysis textbook, and in the chapter on integrating power series it shows $$\int \dfrac{1}{1-(-x)}\mathrm dx=\int\sum_{k=0}^{\infty}(-1)^kx^{k}\mathrm dx=\sum_{k=1}^{\infty}\dfrac{(-1)^{k+1}}{k}x^{k}$$ and therefore $$\ln(1+x)=\sum_{k=1}^{\infty}\dfrac{(-1)^{k+1}}{k}x^{k}$$
It then uses this to say $$\ln(2)=\sum_{k=1}^{\infty}\dfrac{(-1)^{k+1}}{k}$$ This seems specious to me because our prior equality for the geometric series, which served as our starting point for achieving this $\ln(1+x)$ equality, only seemed to be valid on the interval $(-1,1)$, so it seems as if we're missing some justification for why it still holds at $x=1$. Can someone explain to me why this is true?
Best Answer
For $-1<x<1$ the exchange of integration and summation in $$\ln(1+x)=\int_0^x\frac{dx}{1+t}=\int_0^x\sum_{n=0}^\infty(-t)^n\,dx =\sum_{n=0}^\infty\int_0^x(-t)^n\,dt =\sum_{n=0}^\infty\frac{(-1)^nx^{n+1}}{n+1}$$ is valid, say by uniform comvergence.
Abel's theorem states that if $\sum_{n=0}^\infty a_n$ is convergent, then $$\sum_{n=0}^\infty a_n=\lim_{x\to1^-}\sum_{n=0}^\infty a_nx^n.$$ By Abel's theorem, $$\sum_{n=0}^\infty\frac{(-1)^n}{n+1}=\lim_{x\to1^-}\ln(1+x)=\ln2.$$