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I had recently playing around with numbers and found this:

Assume $x$ is a variable, then $$\frac1{(2x)^3}+\frac1{(2x)^5}+\frac1{(2x)^7}+\frac1{(2x)^9}+...=\frac{\frac{1}{(2x)^3}}{1-\frac{1}{(2x)^2}}=\frac12(\frac1{2x-1}+\frac1{2x+1})-\frac1{2x}$$ $$\frac1{(3x)^3}+\frac1{(3x)^5}+\frac1{(3x)^7}+\frac1{(3x)^9}+...=\frac{\frac{1}{(3x)^3}}{1-\frac{1}{(3x)^2}}=\frac12(\frac1{3x-1}+\frac1{3x+1})-\frac1{3x}$$

Or more generally $$\frac1{(nx)^3}+\frac1{(nx)^5}+\frac1{(nx)^7}+\frac1{(nx)^9}+...=\frac{\frac{1}{(nx)^3}}{1-\frac{1}{(nx)^2}}=\frac12(\frac1{nx-1}+\frac1{nx+1})-\frac1{nx}$$ By summing up all cases of $n$ when $n>1$, we get $$\frac{\zeta(3)-1}{x^3}+\frac{\zeta(5)-1}{x^5}+\frac{\zeta(7)-1}{x^7}+\frac{\zeta(9)-1}{x^9}+...=\frac12(\frac{1}{2x-1}+\frac{1}{3x-1}+\frac{1}{4x-1}+...+\frac{1}{2x+1}+\frac{1}{3x+1}+\frac{1}{4x+1}+...-\frac 1x(\frac12+\frac13+\frac14+...)$$

Multiplying both sides by $x^3$ getting $$\zeta(3)-1+\frac{\zeta(5)-1}{x^2}+\frac{\zeta(7)-1}{x^4}+\frac{\zeta(9)-1}{x^6}+...=\frac12(\frac{x^3}{2x-1}+\frac{x^3}{3x-1}+\frac{x^3}{4x-1}+...+\frac{x^3}{2x+1}+\frac{x^3}{3x+1}+\frac{x^3}{4x+1}+...)-x^2(\frac12+\frac13+\frac14+...)$$ Since on the L.H.S. there is $\zeta(3)-1$ as constants, can we say that $\zeta(3)-1$ is the constant part of $$\frac12(\frac{x^3}{2x-1}+\frac{x^3}{3x-1}+\frac{x^3}{4x-1}+...+\frac{x^3}{2x+1}+\frac{x^3}{3x+1}+\frac{x^3}{4x+1}+...)-x^2(\frac12+\frac13+\frac14+...)$$ Or even $$\zeta(3)-1=\lim_{x\rightarrow\infty}\bigg(\frac12(\frac{x^3}{2x-1}+\frac{x^3}{3x-1}+\frac{x^3}{4x-1}+...+\frac{x^3}{2x+1}+\frac{x^3}{3x+1}+\frac{x^3}{4x+1}+...)-x^2(\frac12+\frac13+\frac14+...)\bigg)$$

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1 Answers1

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Yes, though it's not as fancy as you probably hoped. The Riemann zeta function shows up in many double summations. For example, take a function $$ f(x) = \sum_{k=1}^{\infty} g(k) \: x^k $$ then $$ \sum_{n=2}^{\infty} \frac{1}{n} f(\frac{x}{n^s}) = \sum_{k=1}^{\infty} g(k) \:\sum_{n=2}^{\infty} \frac{x^k}{n^{ks+1}} = \sum_{k=1}^{\infty} g(k) \: (\zeta(ks+1)-1) \: x^k\\ $$

Your case is simply $ g(k) = 1, s=2 $, substituting $ 1/x^2 $ for $ x$, multiplying by $x$, then taking the limit, leaving only the first term.

Note: Your right hand side is a modified version of the meromorphic representation of the digamma function.

DecarbonatedOdes
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