Find sum of the series:
$\sum\limits_{n=0}^{\infty}\frac{1}{2n+1}\left ( \frac{1}{2} \right )^{n}$. I just thought that this series is the derivation of the original one, but it is not.
Then I know that: $\sum\limits_{n=0}^{\infty}x^{n}=\frac{1}{1-x}$, so in my case $x=\frac{1}{2}$, but I really do not know, how I should get the fraction.
Can anyone please help me?
Best Answer
Since$$\sum_{n=0}^\infty x^{2n}=\sum_{n=0}^\infty(x^2)^n=\frac1{1-x^2}$$and since$$\left(\sum_{n=0}^\infty\frac{x^{2n+1}}{2n+1}\right)'=\sum_{n=0}^\infty x^{2n}=\frac1{1-x^2}\text{ and }\sum_{n=0}^\infty\frac{0^{2n+1}}{2n+1}=0,$$you have$$\sum_{n=0}^\infty\frac{x^{2n+1}}{2n+1}=\int_0^x\frac{\mathrm dt}{1-t^2}=\operatorname{arctanh}x.$$Therefore$$x\neq0\implies\sum_{n=0}^\infty\frac{x^{2n}}{2n+1}=\frac{\operatorname{arctanh}x}x.$$In particular\begin{align}\sum_{n=0}^\infty\frac1{2n+1}\left(\frac12\right)^n&=\sum_{n=0}^\infty\frac1{2n+1}\left(\frac1{\sqrt2}\right)^{2n}\\&=\frac{\operatorname{arctanh}\left(\frac1{\sqrt2}\right)}{\frac1{\sqrt2}}\\&=\sqrt 2\operatorname{arctanh}\left(\frac1{\sqrt2}\right)\\&=\frac1{\sqrt2}\log\left(\frac{\sqrt2+1}{\sqrt2-1}\right).\end{align}