How do I evaluate $\sum_{k = 1}^{\infty}\big[\frac{(-1)^{k - 1}}{k}\sum_{n = 0}^{\infty}\big\{\frac{1}{k(2^n) + 1}\big\}\big]$?
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 $$\sum^\infty_{k=1}\left(\frac1k-\frac1{k+1}\right)=1.$$
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.
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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.
$$\sum_{k = 1}^{\infty}\Bigg[\frac{(-1)^{k - 1}}{k}\sum_{n = 0}^{\infty}\Bigg\{\frac{1}{k(2^n) + 1}\Bigg\}\Bigg]$$ 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 $k$ $$S_k=\frac{31}{16 k}-\frac{341}{256 k^2}+\frac{4681}{4096 k^3}-\frac{69905}{65536 k^4}+O\left(\frac{1}{k^5}\right)$$