Series of functions to ln

Question

I have the following exercise:

on $[1/2,1]$ study convergence of the derivative of $\sum_{i=0}^\infty \frac{1}{i}(\frac{x-1}{x})^i$

and show that $\sum_{i=0}^\infty \frac{(-1)^{i-1}}{i}$$=ln2$

I already studied convergence of the series $\sum_{i=0}^\infty \frac{1}{i}(\frac{x-1}{x})^i$ and proved that it converges uniformly on every compact of $[1/2,1]$ but I dont know the sum of the series(the sum function), also, I cant see how $ln$ appear in the derivative

Yes but I dont know if Im supposed to use the Taylor series, is there another way? — guest, Oct 06 '15 at 16:25
Possible duplicate of The sum $1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\cdots-(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\cdots)$ does not exist. — Mohan Sharma, Jan 15 '18 at 08:47

score -1 · Accepted Answer · answered Oct 08 '15 at 13:26

Ok I found it, the derivative of the series is equal to $\frac{1}{x²}$$\sum_{i=0}^\infty (\frac{x-1}{x})^{i-1}$ since $(\frac{x-1}{x})^{i-1}$ is in absolute less than 1, it converges (to $x$) and the whole derivative converges to $\frac{1}{x}$.

Since the derivative converges uniformly on the domain, we can take the integral and it should give us an equality with our series, but the integral is $ln(x)$.. we evaluate in $1/2$ and we get $ln2$ equals what we were looking for.

Series of functions to ln

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