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)} $

Question

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)} $

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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:

$$ f(x) = (1+x)\sum_{n=0}^\infty (-1)^n \frac{x^{n+1}}{n+1} - (1-x)\sum_{n=0}^\infty \frac{x^{n+1}}{n+1} \iff$$

$$ f(x) = \sum_{n=0}^\infty \frac{2x^{n+2} + (-1)^n x^{n+1} - x^{n+1} }{n+1} $$

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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?

Answer 1

$$ f(x) = \sum_{n=0}^\infty \frac{2x^{n+2} + (-1)^n x^{n+1} - x^{n+1} }{n+1} $$

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

Answer 2

Summing $$S=\sum_{n=1}^{\infty} \frac{1}{n(2n+1)}=2\sum_{k=1}^{\infty} \left( \frac{1}{2n}-\frac{1}{2n+1}\right)=2 [1/2-1/3+ 1/4-1/5_....]$$ $$\implies S=2(1-\ln 2).$$

Answer 3

Alternative approach for the second part:

$$\begin{eqnarray*} S &=& \sum_{n\geq 1}\frac{1}{n(2n+1)}=2\sum_{n\geq 1}\left(\frac{1}{2n}-\frac{1}{2n+1}\right)=2\sum_{n\geq 1}\int_{0}^{1}\left(x^{2n-1}-x^{2n}\right)\,dx\\&=&2\int_{0}^{1}\sum_{n\geq 1}\left(x^{2n-1}-x^{2n}\right)\,dx = 2\int_{0}^{1}\frac{x-x^2}{1-x^2}\,dx = 2\int_{0}^{1}\frac{x}{1+x}\,dx \\ &=&2\int_{0}^{1}\left(1-\frac{1}{x+1}\right)\,dx = 2\left[x-\log(x+1)\right]_{0}^{1} = \color{red}{2(1-\log 2)}. \end{eqnarray*} $$