Sum to infinity of a non-geometric series

sum of a non geometric infinite series: 1/([2n-1]^2) where "n" is an integer greater or equal to 1.
so the first few terms are : 1 + 1/9 + 1/25 + 1/49 + 1/81 + 1/121...

i know the answer is (Pi^2)/8 but i don't understand how to get to it (in the simplest way possible).

The simplest way I know to do this uses the fact that 1 + 1/4 + 1/9 + 1/16 + ... = π^2/6.

1 + 1/9 + 1/25 + 1/49 + 1/81 + 1/121 + ...
= (1 + 1/4 + 1/9 + 1/16 + ...) - (1/4 + 1/16 + 1/36 + 1/64 + ...)
= (1 + 1/4 + 1/9 + 1/16 + ...) - (1/4) (1 + 1/4 + 1/9 + 1/16 + ...)
= (1 - 1/4) (1 + 1/4 + 1/9 + 1/16 + ...)
= (3/4) (π^2/6)
= π^2/8.
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There are several ways to establish the first fact; here's my favorite way.

Note that sin z = z - z^3/3! + z^5/5! - ...

It can be shown that sin z can be factored (for an infinite product expansion) as
sin z = z(1 - z^2/π^2)(1 - z^2/(2π)^2)(1 - z^2/(3π)^2) ...

As a sanity check, both sides have zeros at all integer multiples of π.
Moreover, the product has the same coefficient for z (namely 1) as the series version for sin z.

Now, simply compare coefficients for z^3:
Series for sin z ==> -1/3! = -1/6.
Product for sin z ==> -1/π^2 - 1/(2π)^2 - /(3π)^2 - ...

Hence, they must be equal:
-1/6 = -1/π^2 - 1/(2π)^2 - /(3π)^2 - ...
==> -1/6 = (-1/π^2) [1 + 1/2^2 + 1/3^2 + ...]
==> π^2/6 = 1 + 1/2^2 + 1/3^2 + ...

I hope this helps!