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

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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, 2 months ago

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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, 2 months ago

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@Nihar Mahajan, you happy now? :3 – Prasun Biswas · 1 year, 11 months ago

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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 · 1 year, 11 months ago

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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 · 1 year, 11 months ago

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