Power series approximation for $\ln((1+x)^{(1+x)}) + \ln((1-x)^{(1-x)})$ to calculate $ \sum_{n=1}^\infty \frac{1}{n(2n+1)} $ Problem
Approximate $f(x) = \ln((1+x)^{(1+x)}) + \ln((1-x)^{(1-x)})$  and then calculate $ \sum_{n=1}^\infty \frac{1}{n(2n+1)} $

My attempt
Let
$$f(x) = \ln((1+x)^{(1+x)}) + \ln((1-x)^{(1-x)}) \iff $$
$$f(x) =  (1+x)\ln(1+x) + (1-x)\ln(1-x) \quad $$
We know that the basic Taylor series for $\ln(1+x)$ is
$$ \ln(1+x) = \sum_{n=0}^\infty (-1)^n \frac{x^{n+1}}{n+1} \quad (1)$$
As far as $\ln(1-x)$ is concerned
$$y(x) = \ln(1-x) \iff y'(x) = \frac{-1}{1-x} = - \sum_{n=0}^\infty x^n \text{ (geometric series)} \iff$$
$$y(x) = \int -\sum_{n=0}^\infty x^n  = - \sum_{n=0}^\infty \frac{x^{n+1}}{n+1} \quad (2)$$
Therefore from $f(x), (1), (2)$ we have:
$$ f(x) = (1+x)\sum_{n=0}^\infty (-1)^n \frac{x^{n+1}}{n+1} - (1-x)\sum_{n=0}^\infty \frac{x^{n+1}}{n+1} \iff$$
$$ f(x) = \sum_{n=0}^\infty \frac{2x^{n+2} + (-1)^n x^{n+1} - x^{n+1} }{n+1} $$

Why I hesitate
It all makes sense to me up to this point. But the exercise has a follow up sub-question that requires to find:
$$ \sum_{n=1}^\infty \frac{1}{n(2n+1)} $$
I am pretty sure that this sum is somehow connected with the previous power series that we've found, but I can't find a way to calculate it, so I assume that I have made a mistake.
Any ideas?
 A: Summing $$S=\sum_{n=1}^{\infty} \frac{1}{n(2n+1)}=2\sum_{k=1}^{\infty} \left( \frac{1}{2n}-\frac{1}{2n+1}\right)=2 [1/2-1/3+ 1/4-1/5_....]$$ $$\implies S=2(1-\ln 2).$$
A: Alternative approach for the second part:
$$\begin{eqnarray*} S &=& \sum_{n\geq 1}\frac{1}{n(2n+1)}=2\sum_{n\geq 1}\left(\frac{1}{2n}-\frac{1}{2n+1}\right)=2\sum_{n\geq 1}\int_{0}^{1}\left(x^{2n-1}-x^{2n}\right)\,dx\\&=&2\int_{0}^{1}\sum_{n\geq 1}\left(x^{2n-1}-x^{2n}\right)\,dx = 2\int_{0}^{1}\frac{x-x^2}{1-x^2}\,dx = 2\int_{0}^{1}\frac{x}{1+x}\,dx \\ &=&2\int_{0}^{1}\left(1-\frac{1}{x+1}\right)\,dx = 2\left[x-\log(x+1)\right]_{0}^{1} = \color{red}{2(1-\log 2)}. \end{eqnarray*} $$
A: $$ f(x) = \sum_{n=0}^\infty \frac{2x^{n+2} + (-1)^n x^{n+1} - x^{n+1} }{n+1} $$
Supposing the above is right. We want to change the $n+2$'s to $n+1$'s. To do this, write, by letting $m+1 = n+2$,
$$ \sum_{n=0}^\infty \frac{2x^{n+2}}{n+1}  = \sum_{m=1}^\infty \frac{2x^{m+1}}{m} = \sum_{n=1}^\infty \frac{2x^{n+1}}{n},$$
where, in the last step, we simply changed the dummy variable $m$ to $n$.
I haven't read it very carefully, but sometimes you can get $2n+1$ in the denominator when you're only summing over odd integers.
SPOILER

 

