Sum of power series $\sum_{n\geq 1} \frac{x^n}{n(2n+1)}$.

Question

I have studied the convergence of the series $\sum_{n\geq 1}\frac{x^n}{n(2n+1)}$ and concluded that its domain of convergence is the interval $D=[-1,1]$. I wish now to find its sum in $D\cap[0,\infty)$. My first approach to the problem has been to do the separation $$ \sum_{n\geq 1}\frac{x^n}{n(2n+1)} = \sum_{n\geq 1}\frac{x^n}{n} -2 \sum_{n\geq 1}\frac{x^n}{2n+1} $$ I know that $\sum_{n\geq 1}\frac{x^n}{n} = -\log(1-x)$, but I don't know how to find the sum of the second summatory. I've tried derivation, integration and changing the beggining term of the series but I haven't reached anything. Thanks in advance for your time and help.

Answer 1

Let $x=t^2$ to make $$\sum_{n= 1}^\infty\frac{x^n}{2n+1}=\sum_{n= 1}^\infty\frac{t^{2n}}{2n+1}=\frac 1t\sum_{n= 1}^\infty\frac{t^{2n+1}}{2n+1}=\frac 1t\left(\sum_{n= 0}^\infty\frac{t^{2n+1}}{2n+1} -t\right)=-1+\frac 1t\sum_{n= 0}^\infty\frac{t^{2n+1}}{2n+1} $$ and now, recognize that $$\sum_{n= 0}^\infty\frac{t^{2n+1}}{2n+1}= \tanh ^{-1}(t)$$

Answer 2

Hint:

Split a known series into it's even and odd terms:

$$\sum_{\color{red}{n=2}}^\infty\frac{x^n}n=\sum_{n=1}^\infty\left(\frac{x^{2n}}{2n}+\frac{x^{2n+1}}{2n+1}\right)$$

$$\sum_{n=2}^\infty\frac{x^n}n=-x+\sum_{n=1}^\infty\frac{x^n}n=\dots$$

$$\sum_{n=1}^\infty\frac{x^{2n}}{2n}=\frac12\sum_{n=1}^\infty\frac{t^n}n=\dots\tag{$t=x^2$}$$

Answer 3

Let's keep things real:

For $0\lt x\lt1$ we have

$$\sum_{n\ge1}{x^n\over2n+1}={1\over\sqrt x}\sum_{n\ge1}{(\sqrt x)^{2n+1}\over2n+1} ={1\over\sqrt x}\left(\sum_{m\ge1}{(\sqrt x)^m\over m}-\sum_{m\ge1}{(\sqrt x)^{2m}\over2m}-1 \right) ={1\over\sqrt x}\left(-\log(1-\sqrt x)+{1\over2}\log(1-x)-1\right)$$

For $-1\le x\lt0$ we have

$$\sum_{n\ge1}{x^n\over2n+1}={1\over\sqrt{|x|}}\sum_{n\ge1}{(-1)^n(\sqrt{|x|})^{2n+1}\over2n+1} ={\arctan(\sqrt{|x|})-\sqrt{|x|}\over\sqrt{|x|}}$$

Finally, for $x=0$ we have

$$\sum_{n\ge1}{x^n\over2n+1}=0$$