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I found this in a not very straightforward way, and it seems like a rather strange-looking identity, but there is probably a simple proof. $$\frac{1}{n^x}+\frac{1}{(n+1)^x}=\frac{H_{{n+1,x}}}{n^xH_{n,x}}+\frac{H_{{n-1,x}}}{(n+1)^xH_{n,x}}$$ Where $H_{n,x}$ is the Generalized Harmonic number.

tyobrien
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    $H_{n,x} = \sum_{k = 1}^n \frac{1}{k^x}$? Then $$\frac{H_{n+1,x}}{n^xH_{n,x}} + \frac{H_{n-1,x}}{(n+1)^xH_{n,x}} = \frac{H_{n,x} + \frac{1}{(n+1)^x}}{n^xH_{n,x}} + \frac{H_{n-1,x}}{(n+1)^xH_{n,x}} = \frac{1}{n^x} + \frac{\frac{1}{n^x}}{(n+1)^xH_{n,x}} + \frac{H_{n-1,x}}{(n+1)^xH_{n,x}} = \frac{1}{n^x} + \frac{\frac{1}{n^x} + H_{n-1,x}}{(n+1)^xH_{n,x}}.$$ – Daniel Fischer Nov 23 '16 at 15:02

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Using the simple computation

$$ \frac{H_{n,x}}{n^x}+\frac{H_{n,x}}{(n+1)^x}=\frac{H_{n+1,x}-(n+1)^{-x}}{n^x}+\frac{H_{n-1,x}+n^{-x}}{(n+1)^x}=\frac{H_{n+1,x}}{n^x}+\frac{H_{n-1,x}}{(n+1)^x}, $$ the result follows immediately.

Aweygan
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