more zeta splitting

\phi=\frac{1}{2}(1+\sqrt{5})

\zeta(\phi)=\sum_{n=1}^{\infty} \frac{1}{n^{\phi}}

therefore, amusingly:

\sum_{n=1}^{\infty} \frac{1}{n^{\phi}} = \sum_{n=1}^{\infty} \frac{1}{n^{\phi}} \left(\frac{n+1}{n+1}\right)

\sum_{n=1}^{\infty} \frac{1}{n^{\phi}} \left(\frac{n+1}{n+1}\right) = \sum_{n=1}^{\infty} \frac{1}{n^{\phi}} \left(\frac{n}{n+1}+\frac{1}{n+1}\right)

\sum_{n=1}^{\infty} \frac{1}{n^{\phi}} \left(\frac{n}{n+1}+\frac{1}{n+1}\right) = \sum_{n=1}^{\infty} \frac{1}{n^{\phi}} \left(\frac{1}{1+n^{-1}}+\frac{1}{1+n}\right)

\sum_{n=1}^{\infty} \frac{1}{n^{\phi}} \left(\frac{1}{1+n^{-1}}+\frac{1}{1+n}\right) = \sum_{n=1}^{\infty} \frac{1}{n^{\phi}+n^{\phi-1}} + \frac{1}{n^{\phi}+n^{\phi+1}}

finally:

\sum_{n=1}^{\infty} \frac{1}{n^{\phi}+n^{\phi-1}} + \frac{1}{n^{\phi}+n^{\phi+1}} = \sum_{n=1}^{\infty} \frac{1}{n^{\phi}+\sqrt[\phi]{n}} + \frac{1}{n^{\phi}+n^{\phi^{2}}}

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