How do I evaluate $\sum_{k = 1}^{\infty}\big[\frac{(-1)^{k - 1}}{k}\sum_{n = 0}^{\infty}\big\{\frac{1}{k(2^n) + 1}\big\}\big]$? 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.
 A: $$\sum_{k = 1}^{\infty}\Bigg[\frac{(-1)^{k - 1}}{k}\sum_{n = 0}^{\infty}\Bigg\{\frac{1}{k(2^n) + 1}\Bigg\}\Bigg]$$
The inner sum
$$S_k=\sum_{n = 0}^{\infty}\Bigg\{\frac{1}{k(2^n) + 1}\Bigg\}=\frac 12+\frac{ \psi _2^{(0)}\left(-\frac{\log \left(-\frac{1}{k}\right)}{\log
   (2)}\right)+ \log \left(-\frac{1}{k}\right)}{ \log (2)} < \frac 2 k$$  So, the infinite summation has an upper bound $\frac {\pi ^2}6$.
Numerically, it seems that the asymptotic value is very close to $1$. For
$$T_p=\sum_{k = 1}^{p}{(-1)^{k - 1}}\frac{S_k}{k}$$
$$\left(
\begin{array}{cc}
 p & T_p \\
 25 & 1.0014973 \\
 50 & 0.9996131 \\
 75  & 1.0001739 \\
 100 & 0.9999017 \\
 125 & 1.0000632 \\
 150 & 0.9999561 \\
 175 & 1.0000323 \\
 200 & 0.9999752
\end{array}
\right)$$
For large values of $k$
$$S_k=\frac{31}{16 k}-\frac{341}{256 k^2}+\frac{4681}{4096 k^3}-\frac{69905}{65536
   k^4}+O\left(\frac{1}{k^5}\right)$$
A: Many years ago, I found a rather simple method of convergence acceleration of alternating series. I wondered: what if a series $a_k$ is not alternating, can I transform it, i.e. find a series $b_k$ so that $$\sum^\infty_{k=1}a_k=\sum^\infty_{k=1}(-1)^{k-1}b_k\tag{1}?$$
If the RHS is absolutely convergent, we can write $$\sum^\infty_{k=1}(-1)^{k-1}b_k=\sum^\infty_{k=1}b_k-2\sum^\infty_{k=1}b_{2k}=\sum^\infty_{k=1}(b_k-2\,b_{2k}).$$ So (1) is satisfied if we choose $b_k$ so that $$a_k=b_k-2\,b_{2k}\tag{2}.$$
Replacing in (2) $k$ by $k\,2^n,$ multiplying by $2^n$ and summing from $n=0$ to $\infty,$ we find $$b_k=\sum^\infty_{n=0}2^n\,a_{k\,2^n}\tag{3},$$ provided $\displaystyle\lim_{n\to\infty}2^n\,b_{k\,2^n}=0.$ Now let $$a_k=\frac1{k(k+1)},$$ i.e.
$$b_k=\sum^\infty_{n=0}2^n\frac1{k\,2^n(k\,2^n+1)}=\frac1k\sum^\infty_{n=0}\frac1{k\,2^n+1}.$$ Then, (1) becomes
$$\sum^\infty_{k=1}\frac1{k(k+1)}=\sum^\infty_{k=1}(-1)^{k-1}\frac1k\sum^\infty_{n=0}\frac1{k\,2^n+1},$$ and the LHS is $$\sum^\infty_{k=1}\left(\frac1k-\frac1{k+1}\right)=1.$$
