So, in my notes it says that the series
$ \sum_{n=0}^{\infty} (\frac{x^{2n+1}}{2n+1}-\frac{x^{n+1}}{2n+2}) $
converges uniformly to $\frac{1}{2}log(x+1)$ on $[-a,a]$ for $0<a<1$ but that at $1$ it converges to $log2$
I think I've been able to prove to myself that it converges uniformly, but I have no idea how to show what it converges to exactly, and on different domains.
How might I approach this.
Thanks for the help.
$$f(x)=\sum_{n=0}^{\infty}x^{2n}=\frac1{1-x^2}$$ $$\int f(x) dx=\sum_{n=0}^{\infty}\frac{x^{2n+1}}{2n+1}=\frac12 \ln{\frac{1+x}{1-x}}$$ $$g(x)=\sum_{n=0}^{\infty}x^{n}=\frac1{1-x}$$ $$\frac12\int g(x) dx=\sum_{n=0}^{\infty}\frac{x^{n+1}}{2n+2}=\frac12\ln {\frac{1}{1-x}}$$ Therefore $$\sum_{n=0}^{\infty}\left(\frac{x^{2n+1}}{2n+1}-\frac{x^{n+1}}{2n+2}\right)=\frac12 \ln{\frac{1+x}{1-x}}-\frac12 \ln{\frac{1}{1-x}}=\frac12\ln(1+x)$$
For $x=1$,
$$S=\lim_{n\to\infty}\sum_{k=0}^{n-1}\left(\frac{1}{2k+1}-\frac{1}{2k+2}\right)=\lim_{n\to\infty}\left(\sum_{k=1}^{2n}\frac{1}{k}-2\sum_{k=1}^{n}\frac{1}{2k}\right)$$ $$=\lim_{n\to\infty}\left(\sum_{k=1}^{2n}\frac1k-\sum_{k=1}^{n}\frac1{k}\right)=\lim_{n\to\infty}\sum_{k=n+1}^{2n}\frac1k$$ $$=\lim_{n\to\infty}\sum_{k=1}^{n}\frac1{n+k}=\lim_{n\to\infty}\sum_{k=1}^{n}\frac{\frac1n}{1+\frac kn}$$ $$=\int_1^2\frac{dx}x=\left[\ln x\right]_1^2=\ln 2$$