Calcule de forma explícita la siguiente suma

$$\begin{align}&\sum_{n=1}^{\infty}\frac{1}{2n+1}\left(\frac{x^2}{2}\right)^n =\\ &\\ &\\ &\frac{x^2}{3·2}+\frac{x^4}{5·2^2}+\frac{x^6}{7·2^3}+\frac{x^8}{9·2^4}+....=\\ &\\ &\frac 12 \left(\frac{x}{2·2^{1/2}}+\frac{(-1)^1x}{2·2^{1/2}}\right)+\frac 12 \left(\frac{x^2}{3·2}+ \frac{(-1)^2x^2}{3·2}\right)+\\ &\\ &\\ &\frac 12 \left(\frac{x^3}{4·2^{3/2}}+ \frac{(-1)^3x^3}{4·2^{3/2}}\right)+\frac 12 \left(\frac{x^4}{5·2^2}+ \frac{(-1)^4x^4}{5·2^2}\right)+...=\\ &\\ &\frac 12 \left(\sum_{n=1}^{\infty}\frac{x^n}{(n+1)2^{n/2}}+\sum_{n=1}^{\infty}\frac{(-1)^nx^n}{(n+1)2^{n/2}}\right) =\\ &\\ &\frac 12 \left(\sum_{n=1}^{\infty}\frac{\left(\frac {x}{\sqrt 2}\right)^n}{n+1}+\sum_{n=1}^{\infty}\frac{(-1)^n\left(\frac {x}{\sqrt 2}\right)^n}{n+1}  \right)=\\ &\\ &\\ &\frac{1}{2}\frac{\sqrt 2}{x}\left(\sum_{n=1}^{\infty}\frac{\left(\frac {x}{\sqrt 2}\right)^{n+1}}{n+1}+\sum_{n=1}^{\infty}\frac{(-1)^n\left(\frac {x}{\sqrt 2}\right)^{n+1}}{n+1}  \right)=\\ &\\ &\\ &\frac{\sqrt 2}{2x}\left(\sum_{n=1}^{\infty}\frac{(-1)^{n+1}\left(-\frac {x}{\sqrt 2}\right)^{n+1}}{n+1}+\sum_{n=1}^{\infty}\frac{(-1)^n\left(\frac {x}{\sqrt 2}\right)^{n+1}}{n+1}  \right)=\\ &\\ &\\ &\frac{\sqrt 2}{2x}\left(-\sum_{n=1}^{\infty}\frac{(-1)^{n}\left(-\frac {x}{\sqrt 2}\right)^{n+1}}{n+1}+\sum_{n=1}^{\infty}\frac{(-1)^n\left(\frac {x}{\sqrt 2}\right)^{n+1}}{n+1}  \right)=\\ &\\ &\text{Se parecen a ln(1+f(x)) pero no tienen}\\ &\text{el término con }x^1\text{por tener exponente n+1}\\ &\\ &\\ &\\ &\frac{\sqrt 2}{2x}\left(-\sum_{n=1}^{\infty}\frac{(-1)^{n-1}\left(-\frac {x}{\sqrt 2}\right)^{n}}{n}-\frac{x}{\sqrt 2}+\sum_{n=1}^{\infty}\frac{(-1)^{n-1}\left(\frac {x}{\sqrt 2}\right)^{n}}{n} -\frac{x}{\sqrt 2} \right)=\\ &\\ &\\ &\frac{\sqrt 2}{2x}\left[-ln\left(1-\frac{x}{\sqrt 2}  \right)  + ln\left(1+\frac{x}{\sqrt 2}  \right)-\frac{2x}{\sqrt 2}\right]=\\ &\\ &\\ &\frac{\sqrt 2}{2x}ln\left( \frac{\sqrt 2+x}{\sqrt 2 -x} \right)-\frac{\sqrt 2}{2x}·\frac{2x}{\sqrt 2} =\\ &\\ &\\ &\frac{\sqrt 2}{2x}ln\left( \frac{\sqrt 2+x}{\sqrt 2 -x} \right)-1\end{align}$$

Y para haber podido usar la serie del logaritmo tenia que ser

|x/sqrt(2)| <1

|x| < sqrt(2)

Converge en (-sqrt(2), sqrt(2))

Y eso es todo.