On the Basel problem

**bkarpuz** · December 28th 2011, 11:53 PM

Dear MHF members,

the problem is related to the so-called Basel sum.

Let
$S:=1+\frac{1}{2^{2}}+\frac{1}{3^{2}}+\cdots$ .
Show that
$1-\frac{1}{2^{2}}-\frac{1}{4^{2}}+\frac{1}{5^{2}}+\frac{1}{7^{2}}-\frac{1}{8^{2}}-\frac{1}{10^{2}}+\frac{1}{11^{2}}+\frac{1}{13^{2}}-\cdot-\cdot+\cdot+\cdots=\frac{4}{9}S$ .

Thanks.
bkarpuz

**sbhatnagar** · December 29th 2011, 01:09 AM

Given that : $\sum_{n=1}^{\infty}\frac{1}{n^2}=S=\frac{\pi^2}{6}$

$\begin{align*} k &=1-\frac{1}{2^2}-\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{7^2}-\cdots \\ &= \left( 1+\frac{1}{3^2}+\frac{1}{5^2}+\frac{1}{7^2} +\cdots \right) - \left( \frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2} +\cdots \right)-\left(\frac{1}{3^2}-\frac{1}{6^2}+\frac{1}{9^2} -\cdots\right)\\ \end{align*}$

Note that :

$[1] \quad \frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2} +\cdots=\sum_{n=1}^{\infty}\frac{1}{(2n)^2}=\frac{ 1}{4}\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{S}{4}$

$[2] \quad 1+\frac{1}{3^2}+\frac{1}{5^2}+\frac{1}{7^2} +\cdots=S-\left( \frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2} +\cdots \right)= S-\frac{S}{4}=\frac{3S}{4}$

$\begin{align*}[3] \quad \frac{1}{3^2}-\frac{1}{6^2}+\frac{1}{9^2} -\cdots &= \left( \frac{1}{3^2}+\frac{1}{9^2}+\frac{1}{15^2}+\cdots \right)-\left( \frac{1}{6^2}+\frac{1}{12^2}+\frac{1}{18^2}+\cdots \right) \\ &=\left( \sum_{n=1}^{\infty}\frac{1}{(3n)^2}-\sum_{n=1}^{\infty}\frac{1}{(6n)^2} \right) - \sum_{n=1}^{\infty}\frac{1}{(6n)^2} \\ &= \frac{S}{9}-\frac{S}{18} \\ &= \frac{S}{18}\end{align*}$

Hence :

$\begin{align*} k &= \left( 1+\frac{1}{3^2}+\frac{1}{5^2}+\frac{1}{7^2} +\cdots \right) - \left( \frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2} +\cdots \right)-\left(\frac{1}{3^2}-\frac{1}{6^2}+\frac{1}{9^2} -\cdots\right) \\ &=\frac{3S}{4}-\frac{S}{4}-\left(\frac{1}{3^2}-\frac{1}{6^2}+\frac{1}{9^2} -\cdots\right) \\ &= \frac{S}{2}-\left(\frac{1}{3^2}-\frac{1}{6^2}+\frac{1}{9^2} -\cdots\right) \\ &=\frac{S}{2}-\frac{S}{18} \\&= \frac{4S}{9}\end{align*}$

Happy New Year!
sbhatnagar

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