I was asked to prove that $f(x)$ is derivative in $x=1/2$ but I found that the radious of convergence is $0$, what did I do wrong? $$ f(x) = \sum_{n=1}^\infty \frac{(-1)^n x^{2n+1}}{2n(2n+1)}$$
I put x outside the summation and declared $t=x^2$ then the radius is equal to limsup of $a_n^{1/n}$ which is $0$.
On
There are two common ways to find the radius of convergence and they are based on the root test and the ratio tests.
The ratio test:
$\sum_{n=1}^{\infty} a_n$ converges if $|\lim_\limits{n\to\infty} \frac {a_n}{a_{n+1}}| < 1$ and diverges if $|\lim_\limits{n\to\infty} \frac {a_n}{a_{n+1}}| > 1$ (and is inconclusive if the ratio equals one.)
The root test.
The series converges if $\lim_\limits{n\to\infty} \sqrt [n]{a_n} < 1$ diverges if that limit is greater than 1 and is inconclusive if it equals 1.
By the ratio test $\frac {a_n}{a_{n+1}} = {\frac {\frac {(-1)^nx^{2n+1}}{2n(2n+1)}}{\frac {(-1)^nx^{2n+3}}{(2n+2)(2n+3)}}} = {-\frac {(2n+2)(2n+3)}{2n(2n+1)x^2}}$
$\lim_\limits{n\to\infty} |-\frac {(2n+2)(2n+3)}{2n(2n+1)x^2}| = \frac {1}{x^2}$ which is less than $1$ when $x< 1$
If $x = 1$ then our our series
$\sum_\limits{n=0}^\infty \frac {(-1)^n}{2n(2n+1)}$ converges.
$|x| \le 1$
The issue is the value of $$\lim_{n\to \infty} \left(2n(2n+1)\right)^{1/n} $$. If this limit converges to some positive number, the radius of convergence is its reciprocal since its limsup is the limit value.
Consider $$\log (2n(2n+1))^{1/n} = \frac{\log(2n(2n+1))}{n} = \frac{\log 2 + \log n + \log (2n+1)}{n}$$
Since $\frac{\log 2}{n}, \frac{\log n}{n}, \frac{\log(2n+1)}{n} \to 0 $ as $n \to \infty$, we have $\lim_{n \to \infty} \log (2n(2n+1))^{1/n} = 0$. Thus we have $$ \log\left( \lim_{n \to \infty} (2n(2n+1))^{1/n}\right) = 0$$ which give $$\lim_{n\to \infty} \left(2n(2n+1)\right)^{1/n} =1 $$
This is the plot of function $f(x) = x^{1/x}$. This function first increases, but eventually starts to to decrease, and converges to $1$ at $\infty$.