A beautiful double sum

Prove that:

$$ \sum_{n=1}^{\infty}\left [ \frac{1}{n} \sum_{m=0}^{\infty} \frac{1}{(2m+1)^{2n}} - \log \left ( 1 + \frac{1}{n} \right ) \right ]$$

Solution

We are basing the solution on the following lemmata:

Lemma 1: It holds that:

$$\sum_{n=1}^{\infty} \left [ \frac{1}{n} - \log \left ( 1  + \frac{1}{n} \right )  \right ] = \gamma $$

where $\gamma$ stands for Euler - Mascheroni constant.

Proof: Well this pretty much follows if we invoke the definition of this particular constant. Thus:

\begin{align*}
\sum_{n=1}^{\infty} \left [ \frac{1}{n} - \log \left ( 1  + \frac{1}{n} \right )  \right ]  &=  \sum_{n=1}^{\infty} \left [ \frac{1}{n} - \log (n+1) + \log n \right ]\\
 &=\lim_{N \rightarrow +\infty} \sum_{n=1}^{N} \left [ \frac{1}{n} - \log (n+1 ) + \log n \right ] \\
 &= \lim_{N \rightarrow +\infty} \left [ \mathcal{H}_N - \log (N+1) \right ]\\
 &= \lim_{N \rightarrow +\infty} \left [ \mathcal{H}_{N+1} - \log (N+1) - \frac{1}{N+1} \right ] \\
 &=\gamma
\end{align*}

ending the proof of Lemma 1.

For our second lemma we will be using  Wallis product to actually reduce a product, namely the product $\displaystyle \prod_{m=1}^{\infty} \left ( 1-\frac{1}{\left ( 2m+1 \right )^{2}} \right ) = \frac{\pi}{4}$.

Lemma 2: It holds that:

$$\prod_{m=1}^{\infty} \left ( 1-\frac{1}{\left ( 2m+1 \right )^{2}} \right ) = \frac{\pi}{4}$$

Proof:  Left to the reader.

 Now on to our exercise. We have successively:

\begin{align*}
\sum_{n=1}^{\infty} \left [ \frac{1}{n} \sum_{m=0}^{\infty} \frac{1}{(2m+1)^{2n}} - \log \left ( 1+ \frac{1}{n} \right ) \right ] &= \sum_{n=1}^{\infty} \left [ \frac{1}{n} + \frac{1}{n} \sum_{m=1}^{\infty} \frac{1}{(2m+1)^{2n}} - \log \left ( 1 + \frac{1}{n} \right ) \right ]\\
 &=  \sum_{n=1}^{\infty} \left [ \frac{1}{n} - \log \left ( 1 + \frac{1}{n} \right ) \right ] + \sum_{n=1}^{\infty} \frac{1}{n} \sum_{m=1}^{\infty} \frac{1}{(2m+1)^{2n}} \\
 &\overset{(1)}{=} \gamma + \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{1}{n \left ( 2m+1 \right )^{2n}} \\
 &= \gamma  - \sum_{m=1}^{\infty} \log \left ( 1 - \frac{1}{(2m+1)^2} \right )\\
 &\overset{(2)}{=}\gamma - \log \left [\prod_{m=1}^{\infty} \left ( 1-\frac{1}{(2m+1)^2} \right )  \right ] \\
 &= \gamma - \log \frac{\pi}{4}
\end{align*}