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!


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

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


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 $(4)$.

  • $\begingroup$ 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
  • 1
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.