Sum of Reciprocal Squares

Calculus Level 3

1 1 2 + 1 2 2 + 1 3 2 + 1 4 2 + ⋯ = π 2 6 \frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\cdots = \frac{\pi^2}{6}

Given the above, find N N that satisfies the equation below.

1 1 2 + 1 3 2 + 1 5 2 + 1 7 2 + ⋯ = π 2 N \frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{5^2}+\frac{1}{7^2}+\cdots = \frac{ \pi^2} { N}


The answer is 8.

Arpon Paul by Arpon Paul , 25, Bangladesh

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1 solution

Arpon Paul
Dec 14, 2016

Given, S = 1 1 2 + 1 2 2 + 1 3 2 + 1 4 2 + . . . ∞ = π 2 6 S = \frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+... \infty = \frac{\pi^2}{6} Let, S 1 = 1 1 2 + 1 3 2 + 1 5 2 + 1 7 2 + . . . ∞ S_1=\frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{5^2}+\frac{1}{7^2}+...\infty and S 2 = 1 2 2 + 1 4 2 + 1 6 2 + 1 8 2 + . . . ∞ S_2=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+... \infty Now, S 2 = 1 4 ( 1 1 2 + 1 2 2 + 1 3 2 + 1 4 2 + . . . ) = S 4 S_2=\frac{1}{4}\left(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+... \right)=\frac{S}{4} Again, S = S 1 + S 2 S=S_1+S_2 ⟹ S = S 1 + S 4 \implies S=S_1+\frac{S}{4} ⟹ S 1 = 3 S 4 = π 2 8 \implies S_1=\frac{3S}{4}=\frac{\pi^2}{8}

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Did the same way, but how can we do it by calculus ? Any hint ??

Bhaskar Pandey - 2 years, 12 months ago

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