Showing that $\frac1{1^2}+\frac1{3^2}+\frac1{5^2}+\cdots =\frac{\pi^2}8$ if $\frac1{1^2}+\frac1{2^2}+\frac1{3^2} +\cdots =\frac{\pi^2}{6}$

Question

Problem:

If
$$\frac{1}{1^2} + \frac{1}{2^2} +\frac{1}{3^2} + \frac{1}{4^2}+\cdots =\frac{\pi^2}{6}$$ show that $$\frac{1}{1^2} +\frac{1}{3^2} + \frac{1}{5^2}+\frac{1}{7^2}\cdots =\frac{\pi^2}{8}$$

This was a question on an exam yesterday. My professor always throws one question in that is above our level and this was the one. I had no idea what to do on the exam. I just wanted to see an answer and the mode of thinking behind said answer.

I am an undergrad student. Sometimes it's helpful to see the work of others to understand a process. I'm an undergrad student in math. The class is history of mathematics.

I've tried many things, multiplying equations, subtracting equations. any help would be appreciated.

Answer 1

Let the required sum be $S$.

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

$$\implies \frac{1}{1^2} +\frac{1}{3^2} + \frac{1}{5^2}+\frac{1}{7^2}+\cdots +\frac{1}{2^2}+\frac{1}{4^2}+\cdots= \frac{\pi^2}{6}$$

$$\implies S+\frac{1}{2^2}\left(\frac{1}{1^2} + \frac{1}{2^2} +\frac{1}{3^2} + \frac{1}{4^2}+\cdots\right)= \frac{\pi^2}{6}$$

$$\implies S+\frac{1}{4} \cdot \frac{\pi^2}{6}=\frac{\pi^2}{6}$$

$$\implies S=\frac{\pi^2}{6}-\frac{\pi^2}{24}=\frac{\pi^2}{8}$$

Hope this helps.

Answer 2

Hints: fill in details (explanations, justifications, etc.)

$$\frac{\pi^2}6=\sum_{n=1}^\infty\frac1{n^2}=\sum_{n=1}^\infty\frac1{(2n)^2}+\sum_{n=1}^\infty\frac1{(2n-1)^2}=\frac14\cdot\frac{\pi^2}6+\sum_{n=1}^\infty\frac1{(2n-1)^2}$$

Answer 3

If you are familiar with Euler products, you can see that

$$1+{1\over3^2}+{1\over5^2}+{1\over7^2}+\cdots=\prod_{p\not=2}\left(1-{1\over p^2}\right)^{-1}=\left(1-{1\over 2^2}\right)\prod_p\left(1-{1\over p^2}\right)^{-1}={3\over4}\left(1+{1\over2^2}+{1\over3^3}+{1\over4^2}+\cdots\right)$$

Answer 4

\begin{align} & \frac 1 {1^2} + \frac 1 {3^2} + \frac 1 {5^2} + \frac 1 {7^2} + \cdots \\[10pt] = {} & \left( \frac 1 {1^2} + \frac 1 {2^2} + \frac 1 {3^3} + \frac 1 {4^2} + \frac 1 {5^2 } + \frac 1 {6^2} + \cdots \right) - \left( \frac 1 {2^2} + \frac 1 {4^2} + \frac 1 {6^2} + \cdots \right) \\[10pt] = {} & \left( \frac 1 {1^2} + \frac 1 {2^2} + \frac 1 {3^3} + \frac 1 {4^2} + \frac 1 {5^2 } + \frac 1 {6^2} + \cdots \right) - \frac 1 {2^2} \left( \frac 1 {1^2} + \frac 1 {2^2} + \frac 1 {3^2} + \cdots \right) \\[10pt] = {} & \frac {\pi^2} 6 - \frac 1 4\cdot\frac{\pi^2} 6. \end{align}