> solution to this series:
>
> 1 + 1/(3^2) + 1/(5^2) + 1/(7^2) + 1/(9^2) + ...

See: http://www.efunda.com/math/seriesofconst/IntRecipSeries.cfm

Or:
Let A = 1 + 1/(3^2) + 1/(5^2) + 1/(7^2) + 1/(9^2) + ...

Using the Riemann's Zeta Function and its 'alternating' form
Zt(s) = Sum_{n=1}^\infty [ 1/n^s ]
Za(s) = Sum_{n=1}^\infty [ (-1)^{n-1} / n^s ]
= [ 1 - 2^{1-s} ] Zt(s)

It is easy to verify that
A = [ Zt(2) + Za(2) ] / 2
= (1/8)pi^2

There are around half-dozen proofs that Zeta(2) = pi^2/6.

========================

Or:
Here is a proof using Fourier series:
 Take the function fx(x) = -x when -π < x < 0 = 0 when 0 < x < π
Notice f(x) is piecewise continuous with period 2π so this means we can find its corresponding Fourier series.

Solve for the coefficients of the Fourier series, and write out the Fourier series for f(x)

f(x) = π/4 - (2/π)*[cos(x) + (1/9)cos(3x) + (1/25)cos(5x) ...] - [sin(x) - (1/2)sin(2x) + (1/3)sin(3x) ...]

Now just evaluate at x=0

You get

0 = π/4 - 2/π * (1 + 1/9 + 1/25 +...)

so π2/8 = (1 + 1/9 + 1/25 + ...) BACK