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Dear Rajesh,
Let f(x) = x^2 This is an even function of x so we know that NONE of the sine terms of the Fourier series can be present. We need only look at the a(i) terms. For all a(n) we get pi.a(n) = INT(-pi to pi)[x^2.cos(nx)dx] When n = 0 this gives a(0) = (2/3)pi^2 When n greater than 0 integration by parts in 3 steps gives a(n) = 4(-1)^n/n^2 The Fourier series is pi^2/3 + 4.SUM(1 to infinity)(-1)^n/n^2 cos(nx) With x=pi, we get pi^2 = pi^2/3 + 4.SUM(-1)^n/n^2 cos(n.pi) and since cos(n.pi)= (-1)^n this produces + s for each term 2.pi^2/3 = 4.SUM[1 + 1/2^2 + 1/3^2 + ..... to infinity] pi^2/6 = 1 + 1/2^2 + 1/3^2 + 1/4^2 + ..... to infinity
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