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Sum of Reciprocals of Odd Squares = pi^2/8 | JEE

JEE Maths question with a full step-by-step solution.

Question

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

, then

\frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{5^2}+\cdots

is equal to

\frac{\pi^2}{8}

correct

\frac{\pi^2}{12}

\frac{\pi^2}{3}

\frac{\pi^2}{2}

Solution

Step 1: The even part factors out

\frac14

\frac{1}{2^2}+\frac{1}{4^2}+\cdots=\frac14\left(\frac{1}{1^2}+\frac{1}{2^2}+\cdots\right)=\frac14\cdot\frac{\pi^2}{6}=\frac{\pi^2}{24}.

Step 2: Odd part

=

total

-

even:

\frac{\pi^2}{6}-\frac{\pi^2}{24}=\frac{4\pi^2-\pi^2}{24}=\frac{\pi^2}{8}.

Correct answer: (1)

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