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# Infinite Series- ISI B.MATH 2006 | Problem - 1

## Problem

If $\sum_{n=1}^{\infty} \frac{1}{n^2} =\frac{{\pi}^2}{6}$ then $\sum_{n=1}^{\infty} \frac{1}{(2n-1)^2}$ is equal to

(A) $\frac{{\pi}^2}{24}$ (B) $\frac{{\pi}^2}{8}$ (C) $\frac{{\pi}^2}{6}$ (D) $\frac{{\pi}^2}{3}$

## Hint

Try to write the summation as sum of square of reciprocal of odd numbers and even numbers and take the advantage of the infinite sum

## Solution

$\sum_{n=1}^{\infty} \frac{1}{n^2} =\frac{{\pi}^2}{6}$

$\Rightarrow \sum_{n=1}^{\infty} \frac{1}{(2n)^2} + \sum_{n=1}^{\infty} \frac{1}{(2n-1)^2}= \frac{{\pi}^2}{6}$

$\Rightarrow \frac{1}{4}\sum_{n=1}^{\infty} \frac{1}{{n^2}} + \sum_{n=1}^{\infty} \frac{1}{(2n-1)^2} = \frac{{\pi}^2}{6}$

we know $\sum_{n=1}^{\infty} \frac{1}{n^2} =\frac{{\pi}^2}{6}$

So from the above equation we get

Hence $\sum_{n=1}^{\infty} \frac{1}{(2n-1)^2} = \frac{{\pi}^2}{6} - \frac{{\pi}^2}{6\cdot4}$

$\Rightarrow \sum_{n=1}^{\infty} \frac{1}{(2n-1)^2} = \frac{{\pi}^2}{8}$

So the correct answer is option B

## Problem

If $\sum_{n=1}^{\infty} \frac{1}{n^2} =\frac{{\pi}^2}{6}$ then $\sum_{n=1}^{\infty} \frac{1}{(2n-1)^2}$ is equal to

(A) $\frac{{\pi}^2}{24}$ (B) $\frac{{\pi}^2}{8}$ (C) $\frac{{\pi}^2}{6}$ (D) $\frac{{\pi}^2}{3}$

## Hint

Try to write the summation as sum of square of reciprocal of odd numbers and even numbers and take the advantage of the infinite sum

## Solution

$\sum_{n=1}^{\infty} \frac{1}{n^2} =\frac{{\pi}^2}{6}$

$\Rightarrow \sum_{n=1}^{\infty} \frac{1}{(2n)^2} + \sum_{n=1}^{\infty} \frac{1}{(2n-1)^2}= \frac{{\pi}^2}{6}$

$\Rightarrow \frac{1}{4}\sum_{n=1}^{\infty} \frac{1}{{n^2}} + \sum_{n=1}^{\infty} \frac{1}{(2n-1)^2} = \frac{{\pi}^2}{6}$

we know $\sum_{n=1}^{\infty} \frac{1}{n^2} =\frac{{\pi}^2}{6}$

So from the above equation we get

Hence $\sum_{n=1}^{\infty} \frac{1}{(2n-1)^2} = \frac{{\pi}^2}{6} - \frac{{\pi}^2}{6\cdot4}$

$\Rightarrow \sum_{n=1}^{\infty} \frac{1}{(2n-1)^2} = \frac{{\pi}^2}{8}$

So the correct answer is option B

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