# 1+(1/2)square+(1/3)square+.....infinity=(pie)square upon 6

Aman Bansal
592 Points
12 years ago

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```

Best Of luck

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Thanks

Aman Bansal