If $\sum\limits_{n=1}^{\infty}\frac{1}{n^{4}}=\frac{\pi^{4}}{90}$
Then find the value of $\sum\limits_{n=1}^{\infty}\frac{1}{(2n-1)^{4}}$
The book I took this problem from makes no mention of Riemann Zeta function at all. In fact, the book had problems only on arithmetic, geometric and harmonic progressions. This is neither one of them. Any hints?

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    $\begingroup$ $\sum_{n=1}^{\infty}\frac{1}{(2n-1)^4}=\frac{15}{16}\zeta(4)=\frac{\pi^4}{96}$. $\endgroup$ – Dietrich Burde Jan 14 '15 at 15:07
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    $\begingroup$ Hint: First find the sum of the other half of the terms. $\endgroup$ – Erick Wong Jan 14 '15 at 15:08
  • $\begingroup$ @Dietrich, Umm... how? $\endgroup$ – AvZ Jan 14 '15 at 15:08
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    $\begingroup$ @AvZ: $\sum_{n=1}^{\infty}\frac{1}{(2n)^4} = \frac{1}{2^4}\sum_{n=1}^{\infty}\frac{1}{n^4}$ $\endgroup$ – Henning Makholm Jan 14 '15 at 15:09

In general we have for all $n\ge 1$ $$ \zeta(n)=\frac{2^n}{2^n-1}\sum_{k=1}^{\infty}\frac{1}{(2k-1)^n} $$ This can be proved as follows \begin{align*} \zeta(n) & = \left(\frac{1}{1^n}+\frac{1}{3^n}+\cdots\right)+\left(\frac{1}{2^n}+\frac{1}{4^n}+\cdots\right) \cr & =\left(\frac{1}{1^n}+\frac{1}{3^n}+\cdots\right)+\frac{1}{2^n}\zeta(n) \cr & = \left(1+\frac{1}{2^n}+\frac{1}{2^{2n}}+\cdots\right)\sum_{k=1}^{\infty} \frac{1}{(2k-1)^n}. \end{align*} Now use the geometric series with $q=\frac{1}{2^n}$.

  • $\begingroup$ where can i study about reimann zeta function? $\endgroup$ – RE60K Jan 14 '15 at 15:26
  • $\begingroup$ See here for references. $\endgroup$ – Dietrich Burde Jan 14 '15 at 15:39
  • $\begingroup$ Nitpick: $k=1$ instead of $k=0$ on the first line. $\endgroup$ – Erick Wong Jan 14 '15 at 17:42
  • $\begingroup$ @ErickWong Thank you, corrected. $\endgroup$ – Dietrich Burde Jan 14 '15 at 18:35

Hint: Break up the sum into even and odd parts. $$\sum_{i=1}^\infty\frac{1}{n^4} = \sum_{i=1}^\infty\frac{1}{(2n-1)^4}+\sum_{i=1}^\infty\frac{1}{(2n)^4}$$ You know the value of the quantity on the LHS, and can easily find the value of $\sum_{i=1}^\infty\frac{1}{(2n)^4}$. Solve for the remaining quantity and you have your answer.


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