Having misread the recent question here as $$\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)(2n+3)}$$ and having solved it, I thought that I would post it as a question instead. It has a rather interesting answer!


Now that we have nice solutions from Jack D'Aurizio and Michael Biro, I'd like to point out that what struck me was the fact that

$$\begin{align} \sum_{n=0}^\infty=\frac \pi 4-\frac 12&=\sum_{n=0}^\infty (-1)^n\frac 1{2n+1}-\frac 1{2^{n+2}}\\ \color{red}{\frac 1{1\color{black}{\cdot 3}}-\frac 1{3\color{black}{\cdot 5}}+\frac 1{5\color{black}{\cdot 7}}-\frac 1{7\color{black}{\cdot 9}}+\cdots }&= \left(\color{red}{\frac 11-\frac 13+\frac15-\frac17+\cdots}\right)-\frac 12\\ &=\left(\color{red}{\frac 11-\frac 13+\frac15-\frac17+\cdots}\right)-\left(\frac 14+\frac 18+\frac 1{16}+\frac 1{32}+\cdots \right) \end{align}$$ Is it possible to reduce the LHS expansion to the RHS expansion directly without first knowing the answer? If so then this would be another solution method.


Combine terms to get

$\sum \frac{(-1)^n}{(2n+1)(2n+3)} = \frac{1}{1 \cdot 3} - \frac{1}{3 \cdot 5} + \dots = \frac{4}{1 \cdot 3 \cdot 5} + \frac{4}{5 \cdot 7 \cdot 9} + \dots = \sum \frac{4}{(4n+1)(4n+3)(4n+5)}$

Partial fractions (and some questionable rearrangement) gives:

$\sum \frac{4}{(4n+1)(4n+3)(4n+5)} = \sum \frac{1}{2(4n+1)} + \frac{1}{2(4n+5)} - \frac{1}{4n+3} $

$= \frac{1}{2}(1 + \frac{1}{5} + \frac{1}{9} + \dots) + \frac{1}{2}(\frac{1}{5} + \frac{1}{9} + \dots) - (\frac{1}{3} + \frac{1}{7} + \dots)$

$=\frac{1}{2} + (-\frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \dots) = \frac{1}{2} + (\arctan(1) - 1 )=\frac{1}{2} + (\frac{\pi}{4} - 1) = \frac{\pi}{4} - \frac{1}{2}$

Rigor is a little suspect, but hey, it works! :)

  • $\begingroup$ Very nice solution! (+1) No calculus invoked. I arrived at this solution as well. Will include further thoughts on this in an edit in the question. $\endgroup$ – hypergeometric Oct 18 '15 at 5:35
  • $\begingroup$ Please see additional comments in the question section. Can you suggest how this might be done? $\endgroup$ – hypergeometric Oct 18 '15 at 10:19

Different question but same technique:

$$ \sum_{n\geq 0}\frac{(-1)^n}{(2n+1)(2n+3)}=\frac{1}{2}\sum_{n\geq 0}(-1)^n\int_{0}^{1}\left(x^{2n}-x^{2n+2}\right)\,dx =\frac{1}{2}\int_{0}^{1}\frac{1-x^2}{1+x^2}\,dx$$ hence:

$$ \sum_{n\geq 0}\frac{(-1)^n}{(2n+1)(2n+3)}=\color{red}{\frac{\pi}{4}-\frac{1}{2}}.$$

  • $\begingroup$ Yes, that's the correct answer (+1). The same technique applies. I'd be interested to see other solutions as well, in particular those without using calculus. $\endgroup$ – hypergeometric Oct 17 '15 at 18:15
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    $\begingroup$ @Jack D'Aurizio, mind you explain what you did there in the end? Just where the summation disappeared. (I must've blinked). $\endgroup$ – Ranc Oct 17 '15 at 18:20
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    $\begingroup$ @Ranc: I switched $\int$ and $\sum$ and computed two geometric series. $\endgroup$ – Jack D'Aurizio Oct 17 '15 at 18:21
  • $\begingroup$ @JackD'Aurizio Just a little followup question on that 1: Usually interchanging the order of $\sum$ and $\int$ is a rather delicate manuever. But It seems there's no need to be cautious in this specific question. Could you explain why's that? $\endgroup$ – Ranc Oct 17 '15 at 18:49
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    $\begingroup$ @Ranc: because the original series is absolutely convergent; the integrals in the middle term behave like $\frac{1}{n^2}$. $\endgroup$ – Jack D'Aurizio Oct 17 '15 at 18:56

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