 # Proving identity \sum[(j+t)^{-1}-j^{-1}]=\sum_{k\geq1}\zeta(k+1)(-t)^k Dashawn Robbins 2022-05-01 Answered
Proving identity
$\sum \left[{\left(j+t\right)}^{-1}-{j}^{-1}\right]=\sum _{k\ge 1}\zeta \left(k+1\right){\left(-t\right)}^{k}$
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Just apply $\sum _{n=0}^{\mathrm{\infty }}{x}^{n}=\frac{1}{1-x}$. For $|t|, we have
${\left(j+t\right)}^{-1}={j}^{-1}{\left(1+\frac{t}{j}\right)}^{-1}$
$={j}^{-1}\sum _{k=0}^{\mathrm{\infty }}{\left(-\frac{t}{j}\right)}^{k}$
Thus
$\sum _{j=1}^{\mathrm{\infty }}\sum _{k=1}^{\mathrm{\infty }}\frac{{\left(-t\right)}^{k}}{{j}^{k+1}}=\sum _{k=1}\sum _{j=1}^{\mathrm{\infty }}\frac{{\left(-t\right)}^{k}}{{j}^{k+1}}$
$=\sum _{k=1}^{\mathrm{\infty }}{\left(-t\right)}^{k}\left(\sum _{j=1}^{\mathrm{\infty }}\frac{1}{{j}^{k+1}}\right)$
$=\sum _{k=1}^{\mathrm{\infty }}{\left(-t\right)}^{k}\zeta \left(k+1\right)$