$\displaystyle\zeta(2) =\frac{\pi^2}{6}!$

Consider the sum, $\displaystyle \zeta(2) = \sum_{r=1}^\infty \frac{1}{r^2}$

$\displaystyle \sum_{r=1}^\infty \frac{1}{r^2} = \sum_{r=0}^\infty \frac{1}{(2r+1)^2} + \sum_{r=1}^\infty \frac{1}{(2r)^2}$

$\displaystyle \sum_{r=1}^\infty \frac{1}{r^2} = \sum_{r=0}^\infty \frac{1}{(2r+1)^2} + \frac{1}{4}\bigg(\sum_{r=1}^\infty \frac{1}{r^2}\bigg)$

$\Rightarrow \displaystyle \sum_{r=0}^\infty \frac{1}{(2r+1)^2} = \frac{3}{4} \bigg(\sum_{r=1}^\infty \frac{1}{r^2}\bigg)$

$\Rightarrow \displaystyle \sum_{r=0}^\infty \frac{1}{(2r+1)^2} = \frac{3}{4} \times\zeta(2)$

$\Rightarrow \displaystyle \sum_{r=0}^\infty \frac{1}{(2r+1)^2} = \frac{3}{4} \times\frac{\pi^2}{6}$

$\Rightarrow \displaystyle \sum_{r=0}^\infty \frac{1}{(2r+1)^2} = \frac{\pi^2}{8} = \boxed{1.2337}$

I have a slightly harder generalization of this sum. Show that all non-trivial zeros of the complex interpolation of a function, say $\zeta(s)=\sum_{n=1}^\infty\frac1{n^s}$ , occur where $\Re(s)=\frac12$ .

seriesSum(1/(r^2) ,r,1,infinity) = pi^2 /6. = 1+1/4+1/9+1/16+1/25+1/36+1/49+...... Since 2r+1 is odd, seriesSum(1/(2r+1)^2 ,r,0, infinity) = 1+1/9+1/25+1/49+....... = pi^2/6 - 1/4-1/16-1/36-1/64-......= = pi^2 /6 -seriesSum(1/(2r)^2 ,r,1,infinity) = pi^2 /6 -1/4 * seriesSum(1/r^2 ,r,1,infinity) = pi^2 /6 - pi^2 /24 = pi^2 /8 =1.2337005501362.