A cute limit $\lim_{m\to\infty}\left(\left(\sum_{n=1}^{m}\frac{1}{n}\sum_{k=1}^{n-1}\frac{(-1)^k}{k}\right)+\log(2)H_m\right)$

Question

I'm sure that for many of you this is a limit pretty easy to compute, but my concern here is a bit different, and I'd like to know if I can nicely compute it without using special functions. Do you have in mind such ways?

$$\lim_{m\to\infty}\left(\left(\sum_{n=1}^{m}\frac{1}{n}\sum_{k=1}^{n-1}\frac{(-1)^k}{k}\right)+\log(2)H_m\right)$$

The limit is $\displaystyle;\frac{\pi^2}{12} - \frac{\log^2 2}{2};$. All you need is taylor expansion of $\log(1+t)$ and the well known identity $\sum\limits_{k=1}^\infty \frac{1}{k^2} = \frac{\pi^2}{6}$. — achille hui, Sep 30 '14 at 13:55

score 12 · Accepted Answer · edited Apr 13 '17 at 12:20

12

$$ \begin{align} \lim_{m\to\infty}\left(\sum_{n=1}^m\sum_{k=1}^{n-1}\frac{(-1)^k}{nk}+H_m\log(2)\right) &=\lim_{m\to\infty}\sum_{n=1}^m\sum_{k=n}^\infty\frac{(-1)^{k-1}}{nk}\\ &=\sum_{n=1}^\infty\sum_{k=n}^\infty\frac{(-1)^{k-1}}{nk}\\ &=\sum_{k=1}^\infty\sum_{n=1}^k\frac{(-1)^{k-1}}{nk}\\ &=\sum_{k=1}^\infty(-1)^{k-1}\frac{H_k}k\\ &=\frac12\zeta(2)-\frac12\log(2)^2 \end{align} $$ the last step is from this answer where only series manipulations are used, no special functions.

edited Apr 13 '17 at 12:20

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answered Sep 30 '14 at 14:55

robjohn

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@robjohn I see you are a legend at summations. Can you tell me from where I can learn all the tricks you use? – Aditya Kumar Mar 13 '16 at 16:29
@AdityaKumar: experience from computing many sums. – robjohn Mar 13 '16 at 18:04

A cute limit $\lim_{m\to\infty}\left(\left(\sum_{n=1}^{m}\frac{1}{n}\sum_{k=1}^{n-1}\frac{(-1)^k}{k}\right)+\log(2)H_m\right)$

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