TITLE: How do I evaluate $\sum_{k = 1}^{\infty}\big[\frac{(-1)^{k - 1}}{k}\sum_{n = 0}^{\infty}\big\{\frac{1}{k(2^n) + 1}\big\}\big]$?
QUESTION [9 upvotes]: Evaluate the following summation: $$\sum_{k = 1}^{\infty}\Bigg[\frac{(-1)^{k - 1}}{k}\sum_{n = 0}^{\infty}\Bigg\{\frac{1}{k(2^n) + 1}\Bigg\}\Bigg]$$
My attempts were to telescope by introducing the $k$ inside the inner summation and using partial fractions, but to no avail. I had noticed a logarithmic series across the summation but it leaves a more complicated sum. There is no algebraic identity I am aware of that simplifies the denominator, and I am unable to see a possible binomial series.

REPLY [8 votes]: Remarks: 1. I made some mistakes. Fortunately, I overcame those mistakes.

Thank @Martin Argerami for the valuable comment. I rewrote the solution by adding the reason of the interchanging order of summation.

$\phantom{2}$
Denote
$$S = \sum_{k=1}^\infty\left[\frac{(-1)^{k-1}}{k}\sum_{n=0}^\infty \frac{1}{k 2^n + 1}\right].$$
Since $\sum_{n=0}^\infty \frac{1}{k 2^n + 1} \le \sum_{n=0}^\infty \frac{1}{k 2^n} = \frac{2}{k}$,
$\frac{1}{k}\sum_{n=0}^\infty \frac{1}{k 2^n + 1}
\le \frac{2}{k^2}$ and $\sum_{k=1}^\infty \frac{2}{k^2} = \frac{\pi^2}{3}$, we know that $S$ converges absolutely. We can interchange the order of summation to get
\begin{align}
S &= \sum_{n=0}^\infty \sum_{k=1}^\infty \frac{(-1)^{k-1}}{k}\frac{1}{k 2^n + 1}\\
&= \sum_{k=1}^\infty \frac{(-1)^{k-1}}{k}\frac{1}{k + 1} + \sum_{n=1}^\infty \sum_{k=1}^\infty \frac{(-1)^{k-1}}{k}\frac{1}{k 2^n + 1}\\
&= \sum_{k=1}^\infty \frac{(-1)^{k-1}}{k} - \sum_{k=1}^\infty \frac{(-1)^{k-1}}{k+1}
 + \sum_{n=1}^\infty \sum_{k=1}^\infty \frac{(-1)^{k-1}}{k}\frac{1}{k 2^n + 1}\\
&= 2\ln 2 - 1 + \sum_{n=1}^\infty \sum_{k=1}^\infty \frac{(-1)^{k-1}}{k}\frac{1}{k 2^n + 1}
\end{align}
where we have used $\sum_{k=1}^\infty \frac{(-1)^{k-1}}{k} = \ln 2$
and $- \sum_{k=1}^\infty \frac{(-1)^{k-1}}{k+1} = \sum_{k=1}^\infty \frac{(-1)^{k}}{k+1}
= \sum_{k=0}^\infty \frac{(-1)^{k}}{k+1} - 1 = \ln 2 - 1$.
Denote $S_1 = \sum_{n=1}^\infty \sum_{k=1}^\infty \frac{(-1)^{k-1}}{k}\frac{1}{k 2^n + 1}$.
By using $\sum_{j=0}^\infty \frac{(-1)^j}{(k 2^n)^{j+1}} = \frac{1}{k 2^n + 1}$, we have
$$S_1 = \sum_{n=1}^\infty \sum_{k=1}^\infty \sum_{j=0}^\infty \frac{(-1)^{k-1}}{k}\frac{(-1)^j}{(k 2^n)^{j+1}}.$$
Since $\sum_{j=0}^\infty \frac{1}{k}\frac{1}{(k 2^n)^{j+1}} = \frac{1}{k}\frac{1}{k 2^n - 1}
\le \frac{1}{k} \frac{1}{k 2^{n-1}}$ and $\sum_{n=1}^\infty \sum_{k=1}^\infty \frac{1}{k} \frac{1}{k 2^{n-1}}
= \frac{\pi^2}{3}$, we know that $S_1$ converges absolutely. We can interchange the order of summation to get
\begin{align}
S_1 &= \sum_{j=0}^\infty \sum_{n=1}^\infty \sum_{k=1}^\infty  \frac{(-1)^{k-1}}{k}\frac{(-1)^j}{(k 2^n)^{j+1}}\\
&= \sum_{j=0}^\infty \left[(-1)^j \left(\sum_{n=1}^\infty \frac{1}{2^{n(j+1)}}\right)
\left( \sum_{k=1}^\infty  \frac{(-1)^{k-1}}{k^{j+2}}\right)\right]\\
&= \sum_{j=0}^\infty \left[(-1)^j \frac{1}{2^{j+1}-1}
\cdot (1-2^{-j-1})\sum_{m=1}^\infty \frac{1}{m^{j+2}}\right]\\
&= \sum_{j=0}^\infty \left[\frac{(-1)^j}{2^{j+1}} \sum_{m=1}^\infty \frac{1}{m^{j+2}}\right]
\end{align}
where $\sum_{k=1}^\infty  \frac{(-1)^{k-1}}{k^{j+2}} = (1-2^{-j-1})\sum_{m=1}^\infty \frac{1}{m^{j+2}}$ follows from
\begin{align}
&1 - \frac{1}{2^{j+2}} + \frac{1}{3^{j+2}} - \frac{1}{4^{j+2}} + \frac{1}{5^{j+2}} -  \frac{1}{6^{j+2}} \cdots \\
=\ & \left(1 + \frac{1}{2^{j+2}} + \frac{1}{3^{j+2}} + \frac{1}{4^{j+2}} + \cdots\right)
- 2\left(\frac{1}{2^{j+2}} + \frac{1}{4^{j+2}} + \frac{1}{6^{j+2}} + \cdots\right)\\
=\ & \left(1 + \frac{1}{2^{j+2}} + \frac{1}{3^{j+2}} + \frac{1}{4^{j+2}} + \cdots\right)
- \frac{1}{2^{j+1}}\left(1 + \frac{1}{2^{j+2}} + \frac{1}{3^{j+2}} + \frac{1}{4^{j+2}} + \cdots\right)\\
=\ & (1 - 2^{-j-1})\left(1 + \frac{1}{2^{j+2}} + \frac{1}{3^{j+2}} + \frac{1}{4^{j+2}} + \cdots\right).
\end{align}
Since $\frac{1}{2^{j+1}} \sum_{m=1}^\infty \frac{1}{m^{j+2}}
\le \frac{1}{2^{j+1}} \sum_{m=1}^\infty \frac{1}{m^{2}} = \frac{\pi^2}{6} \frac{1}{2^{j+1}} $
and $\sum_{j=0}^\infty \frac{\pi^2}{6} \frac{1}{2^{j+1}} = \frac{\pi^2}{6} $,
we know that $\sum_{j=0}^\infty \left[\frac{(-1)^j}{2^{j+1}} \sum_{m=1}^\infty \frac{1}{m^{j+2}}\right]$
converges absolutely. Thus, we can interchange the order of summation to get
\begin{align}
S_1 &= \sum_{m=1}^\infty \sum_{j=0}^\infty \frac{(-1)^j}{2^{j+1}}  \frac{1}{m^{j+2}}\\
&= \sum_{m=1}^\infty \frac{1}{m}\sum_{j=0}^\infty \frac{(-1)^j}{(2m)^{j+1}}\\
&= \sum_{m=1}^\infty \frac{1}{m(2m+1)}\\
&= 2 \sum_{m=1}^\infty \left(\frac{1}{2m} - \frac{1}{2m+1}\right)\\
&= 2 \left(1 - \sum_{m=1}^\infty \frac{(-1)^{m-1}}{m}\right)\\
&= 2(1 - \ln 2)
\end{align}
where we have used $\sum_{m=1}^\infty \frac{(-1)^{m-1}}{m} = \ln 2$.
Finally, we have $S = 2\ln 2 - 1 + S_1 = 1$. We are done.