For the Fourier series in Problem 1, deduce a series for 4π at the point where x=2π
f(x)=2a0+Σansin2nπx+Σbncos2nπx
but bn = 0
Hence, f(x)=2a0+Σansin2nπx
When 2π, f(x) = 2, hence;
2=π8sin2π+3π8sin23π+5π8sin25π+...
i.e
2=π8(1−31+51−71+...)
4π=1−31+51−71+...
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