Prove the following:

\[\frac{\zeta(2) }{2}+\frac{\zeta (4)}{2^3}+\frac{\zeta (6)}{2^5}+\frac{\zeta (8)}{2^7}+\cdots=1\]

Note: There's a very elegant proof to this which doesn't use integral calculus and uses only changing of summation order and telescoping sum. Can you find it?

This note was inspired by a friend who wanted me to post more problems for the community.

## Comments

Sort by:

TopNewest@Prasun Biswas consider \[\zeta(s)=\sum_{n=1}^\infty \dfrac{1}{n^s}\] the required summation is \[2\sum_{s=1}^\infty \sum_{n=1}^\infty \dfrac{1}{(2n)^{2s}}=2\sum_{n=1}^\infty \dfrac{1}{(2n)^2-1}\\ =\sum_{n=1}^\infty\left( \dfrac{1}{2n-1}-\dfrac{1}{2n+1}\right) = 1\] – Aareyan Manzoor · 1 year, 5 months ago

Log in to reply

You should, however, elaborate your solution to mention all the non-trivial steps taken in your solution. For example, don't just start with the double sum. It would be much better to show how the double sum form is obtained.

\[S=\sum_{s=1}^\infty\frac{\zeta(2s)}{2^{2s-1}}=2\sum_{s=1}^\infty\frac{\zeta(2s)}{2^{2s}}=2\sum_{s=1}^\infty\frac{\displaystyle\sum_{n=1}^\infty\dfrac{1}{n^{2s}}}{2^{2s}}=2\sum_{s=1}^\infty\sum_{n=1}^\infty\frac{1}{(2n)^{2s}}\]

Also, a brief mention of the interchange of summation order and why is it allowed here would be nice. – Prasun Biswas · 1 year, 5 months ago

Log in to reply

@Nihar Mahajan, you happy now? :3 – Prasun Biswas · 2 years, 2 months ago

Log in to reply

The answer is kind off obvious after you mentioned manipulating sums, but wouldn't you first have to show that (a) the sum is finite and has a limit of 1 and (b) that rearrangement is allowed. IIRC, you can't just rearrange terms. – Siddhartha Srivastava · 2 years, 2 months ago

Log in to reply

b) Rearrangement is allowed here by a special case of Tonelli's theorem which you'd be able to identify later since the "stuff" you'd be summing will be non-negative for all values through which \(i,j\) cycles where \(i,j\) are the indexes of the double sum you need to form. – Prasun Biswas · 2 years, 2 months ago

Log in to reply