> 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 |

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 π^{2}/8 = (1 + 1/9 + 1/25 + ...)

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