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Question

Find the sum of the series:
∑n=1∞1n2  =  1+14+19+116+⋯\sum_{n=1}^{\infty} \frac{1}{n^2} \;=\; 1 + \frac{1}{4} + \frac{1}{9} + \frac{1}{16} + \cdots

Solution

π26\dfrac{\pi^2}{6}
1
Consider f(x)=x2f(x) = x^2 on (−π, π)(-\pi,\,\pi). Its Fourier series expansion:
f(x)=a02+∑n=1∞(ancos⁡nx+bnsin⁡nx)f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty}\bigl(a_n \cos nx + b_n \sin nx\bigr)
2
Since f(x)=x2f(x) = x^2 is even, all bn=0b_n = 0. Computing the coefficients:
a0=2π23,an=4(−1)nn2a_0 = \frac{2\pi^2}{3}, \qquad a_n = \frac{4(-1)^n}{n^2}
3
Substituting gives the Fourier series for x2x^2:
x2=π23+4∑n=1∞(−1)ncos⁡nxn2x^2 = \frac{\pi^2}{3} + 4\sum_{n=1}^{\infty} (-1)^n \frac{\cos nx}{n^2}
4
Substitute x=πx = \pi. Since cos⁡(nπ)=(−1)n\cos(n\pi) = (-1)^n:
π2=π23+4∑n=1∞(−1)n⋅(−1)nn2=π23+4∑n=1∞1n2\pi^2 = \frac{\pi^2}{3} + 4\sum_{n=1}^{\infty} (-1)^n \cdot \frac{(-1)^n}{n^2} = \frac{\pi^2}{3} + 4\sum_{n=1}^{\infty} \frac{1}{n^2}
5
Isolate the sum:
4∑n=1∞1n2=π2−π23=2π234\sum_{n=1}^{\infty} \frac{1}{n^2} = \pi^2 - \frac{\pi^2}{3} = \frac{2\pi^2}{3}
6
Divide both sides by 44:
  ∑n=1∞1n2=π26  \boxed{\;\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}\;}

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