It is very easy to show that the series $$\frac{1-1/2}{1\times2} - \frac{1-1/2+1/3}{2\times3} + \frac{1-1/2+1/3-1/4}{3\times4} - ...$$ i.e. $$\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n(n+1)}[1-\frac{1}{2} + \frac{1}{3} - ...+ \frac {(-1)^{n}}{n+1}]$$ is convergent. Can one find its exact value? Or is it unreasonable to hope for such a thing? Thank you for your answers.

  • $\begingroup$ The first thing that came to mind is that you might need Cesaro summability or something (cf. exercise 12-37 in Apostol, Mathematical Analysis), but I'm not sure that gets you where you where you want to go. It seems reasonable that this might be possible. Have you tried Hardy's Divergent Series? There may be a simple answer (one way or the other), but I'm not seeing it. $\endgroup$ – Chris Leary Sep 14 '11 at 17:24

The interior sum is equal to $\int_0^1\frac{1-(-x)^{n+1}}{1+x}dx$ and $$ \sum _{n=1}^{\infty } \frac{(-1)^{n+1} \left(1-(-x)^{n+1}\right)}{n (n+1) (x+1)}= \frac{(x-1) \log (1-x)-x+\log (4)-1}{1+x}. $$ So the answer is eqaul to $$ \int_0^1 \frac{(x-1) \log (1-x)-x+\log (4)-1}{1+x}dx=\frac{\pi ^2}{6}+\log ^2(2)-2. $$

  • $\begingroup$ What do you mean by interior sum? $\endgroup$ – Michael Albanese Sep 4 '12 at 10:24



we have

$$\begin{aligned} \sum_{n=1}^N\sum_{k=1}^{n+1}\frac{(-1)^{n+k}}{kn(n+1)} &=\sum_{n=1}^N\sum_{k=1}^{n+1}\frac{(-1)^{n+k}}{kn}+\sum_{n=2}^{N+1}\sum_{k=1}^n\frac{(-1)^{n+k}}{kn}\\\\ &=2\sum_{n=1}^N\sum_{k=1}^n\frac{(-1)^{n+k}}{kn}-\sum_{n=1}^N\frac1{n(n+1)}-1+\sum_{k=1}^{N+1}\frac{(-1)^{N+1+k}}{k(N+1)}\\\\ &=\sum_{n,k=1}^N\frac{(-1)^{n+k}}{kn}+\sum_{n=1}^N\frac1{n^2}-2+\frac1{N+1}+\frac{(-1)^{N+1}}{N+1}\sum_{k=1}^{N+1}\frac{(-1)^k}k\\\\ &=\left(\sum_{n=1}^N\frac{(-1)^{n}}n\right)^2+\sum_{n=1}^N\frac1{n^2}-2+\frac1{N+1}+\frac{(-1)^{N+1}}{N+1}\sum_{k=1}^{N+1}\frac{(-1)^k}k. \end{aligned}$$

As $\sum_{n=1}^\infty\frac{(-1)^n}n=-\log2$ and $\sum_{n=1}^\infty\frac1{n^2}=\frac{\pi^2}6$, this implies


There may be a numerical error somewhere, but in principle the method should work.

  • 1
    $\begingroup$ This can't be right as stated because it's negative. But your method must work because I get $(\log 2)^2 + \pi^2/6 - 2$ by a different approach. $\endgroup$ – Noam D. Elkies Sep 14 '11 at 18:34
  • 1
    $\begingroup$ ...which I see that Andrew just posted. $\endgroup$ – Noam D. Elkies Sep 14 '11 at 18:35

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