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Evaluate: $$\sum_{n=1}^{\infty} {\left(\frac{(-1)^n}{n}\sum_{k=1}^n {\left(\frac{(-1)^k}{k}\right)}\right)}$$ It looks like the series for $\ln(2)$ 'embedded in itself', so my guess for the value is $\ln^2(2)$. Unfortunately, this is not correct, as an estimate for the sum is $1.06269$, much greater than $\ln^2(2)\approx 0.48045$.

Any ideas? Thanks!

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Let $S$ be given by

$$S=\sum_{n=1}^\infty \frac{(-1)^n}{n}\sum_{k=1}^n \frac{(-1)^k}{k}\tag1$$


We can interchange the order of summation in $(1)$ and express $S$ as

$$S=\sum_{k=1}^\infty \frac{(-1)^k}{k}\sum_{n=k}^\infty \frac{(-1)^n}{n} \tag2$$

whereupon switching "dummy" summation indices in $(2)$ yields

$$\begin{align}S&=\sum_{n=1}^\infty \frac{(-1)^n}{n}\sum_{k=n}^\infty \frac{(-1)^k}{k}\\\\ &=\sum_{n=1}^\infty \frac{(-1)^n}{n}\left(\sum_{k=1}^\infty \frac{(-1)^k}{k}-\sum_{k=1}^{n-1} \frac{(-1)^k}{k}\right)\\\\ &=\log^2(2)-\sum_{n=1}^\infty \frac{(-1)^n}{n} \sum_{k=1}^{n-1} \frac{(-1)^k}{k}\\\\ &=\log^2(2)-S+\sum_{n=1}^\infty\frac{1}{n^2}\tag3 \end{align}$$


Using $\sum_{n=1}^\infty\frac{1}{n^2}=\frac{\pi^2}{6}$ in $(3)$ and solving for $S$ we find

$$\bbox[5px,border:2px solid #C0A000]{S=\frac12 \log^2(2)+\frac{\pi^2}{12}}\tag 4$$


NOTE:

The result in $(4)$ is not at all surprising given the symmetry. We can write in general

$$\begin{align} \left(\sum_{n=1}^\infty a_n \right)^2&=\sum_{n=1}^\infty a_n \sum_{k=1}^\infty a_k\\\\ &=2\sum_{n=1}^\infty a_n \sum_{k=1}^{n} a_k-\sum_{n=1}^\infty a_n^2 \end{align}$$

which becomes upon rearranging

$$\sum_{n=1}^\infty a_n \sum_{k=1}^{n} a_k=\frac12 \left(\sum_{n=1}^\infty a_n \right)^2+\frac12 \sum_{n=1}^\infty a_n^2$$

Finally, setting $a_n=\frac{(-1)^n}{n}$, we recover the result in $(4)$.

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  • $\begingroup$ could you explain how the "switch" of summation works? $\endgroup$ – DeepSea Jun 14 '17 at 3:49
  • $\begingroup$ @DeepSea $$\sum_{n=1}^N a_n \sum_{k=1}^n a_k=\sum_{k=1}^N a_k \sum_{n=k}^N a_n$$The entire development proceeds with our taking $N\to \infty$ as a last step. $\endgroup$ – Mark Viola Jun 14 '17 at 3:52
  • $\begingroup$ Wow, your answer is perfect! The way the sum from 1 to n was transformed into the sum from n to infinity was really nice, and I appreciated the note that generalized the problem. Thanks! $\endgroup$ – Ant Jun 14 '17 at 4:38
  • $\begingroup$ The series is only conditionally convergent, so it's not totally obvious that changing the order of summation is valid. There is something to check. $\endgroup$ – Julian Rosen Jun 14 '17 at 5:27
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    $\begingroup$ @i707107 We now have the solution supplemented with comments, which will give future users a complete and comprehensive way forward. $\endgroup$ – Mark Viola Jun 14 '17 at 18:23

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