Proof why $\frac{1}{n^x}+\frac{1}{(n+1)^x}=\frac{H_{{n+1,x}}}{n^xH_{n,x}}+\frac{H_{{n-1,x}}}{(n+1)^xH_{n,x}}$

Question

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.

Answer 1

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.