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Are there other methods to evaluate the following series? $$\frac{1^{-4}+2^{-4}+3^{-4}+4^{-4}+\cdots}{1^{-4}+3^{-4}+5^{-4}+7^{-4}+\cdots}$$

My attempt is as follows,

\begin{align} \frac{1^{-4}+2^{-4}+3^{-4}+4^{-4}+\cdots}{1^{-4}+3^{-4}+5^{-4}+7^{-4}+\cdots} &=\frac{x}{y} \\ \frac{1^{-4}+3^{-4}+\cdots+2^{-4}+4^{-4}+\cdots}{1^{-4}+3^{-4}+5^{-4}+7^{-4}+\cdots} &=\frac{x}{y} \\ \frac{y+2^{-4}(1^{-4}+2^{-4}+3^{-4}+4^{-4}+\cdots)}{y} &=\frac{x}{y} \\ \frac{y+2^{-4}x}{y} &=\frac{x}{y} \\ \frac{x}{y} &= \frac{1}{1-2^{-4}} \end{align}

Greg Martin
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kiss my armpit
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    That's about as slick as you can hope, and it's a solid method. People used similar reasoning to figure out how to extend the Riemann $\zeta$-function to a larger domain with that trick. – Adam Hughes Aug 27 '14 at 17:34
  • Certainly, better to use $\sum$-notation, but otherwise, the best approach. – Thomas Andrews Aug 27 '14 at 17:37

1 Answers1

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Your proof is fine.

The sigma notation can make some things cleaner. The below is the same argument, just written with different notation: $$x=\sum_{n=1}^\infty \frac{1}{n^4}\\y=\sum_{n=1}^\infty \frac{1}{(2n-1)^4}$$

Then $$x=\sum_{n=1}^{\infty} \left(\frac{1}{(2n-1)^4}+\frac{1}{(2n)^4}\right)=y+\frac{1}{2^4}x$$

and thus $$\frac{x}{y} = 1+\frac{1}{2^4}\frac{x}{y}\implies\\\frac{x}{y}=\frac{1}{1-\frac{1}{2^4}}$$

I suppose that the other approach is to note that every $n$ can be written uniquely as $2^k(2m+1)$. So:

$$\begin{align}x = \sum_{n=1}^\infty \frac{1}{n^4} &= \sum_{k=0}^\infty\sum_{m=1}^{\infty} \frac{1}{(2^k(2m+1))^4}\\ &=\left(\sum_{k=0}^\infty \frac{1}{2^{4k}}\right)\left(\sum_{m=0}^\infty \frac{1}{(2m+1)^4}\right)\\ &=\frac{1}{1-\frac{1}{2^4}}y \end{align}$$

Thomas Andrews
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