Proving identity $\sum_{j\geq 1}[(j+t)^{-1}-j^{-1}]=\sum_{k\geq 1}\zeta (k+1)(-t)^{k}$

Question

Motivation: In S.J. Patterson's An introduction to the theory of the Riemann Zeta-Function it is proved (p.132) that

$\displaystyle -\Gamma ^{\prime }(t)/\Gamma (t)=\gamma +t^{-1}+\underset{j\geq 1}{\sum }[(j+t)^{-1}-j^{-1}]$.

In Exercise A4.1 (p.135) one is asked to show that if $|t|<1$

$\displaystyle -\Gamma ^{\prime }(t)/\Gamma (t)=\gamma +t^{-1}+\underset{k\geq 1}{\sum }\zeta (k+1)(-t)^{k}$.

Question: How does one prove that

$\displaystyle\underset{j\geq 1}{\sum }[(j+t)^{-1}-j^{-1}]=\underset{k\geq 1}{\sum }\zeta (k+1)(-t)^{k}\qquad $for $|t|<1$?

Answer 1

Just apply $\sum_{n=0}^\infty x^n = \frac1{1-x}$. For $|t| < \min j = 1$, we have

$$ (j + t)^{-1} = j^{-1} \left( 1 + \frac tj \right)^{-1} = j^{-1} \sum_{k=0}^\infty \left( -\frac tj\right)^k, $$

thus the LHS is

$$ \sum_{j=1}^\infty \sum_{k=1}^\infty \frac{(-t)^k}{j^{k+1}} = \sum_{k=1}^\infty \sum_{j=1}^\infty \frac{(-t)^k}{j^{k+1}} = \sum_{k=1}^\infty (-t)^k \left( \sum_{j=1}^\infty \frac1{j^{k+1}} \right) = \sum_{k=1}^\infty (-t)^k \zeta(k+1) $$

Answer 2

Use $$ \frac{1}{j + t} = \frac{1}{j} \frac{1}{1 + \frac{t}{j}} = \frac{1}{j} \sum_{k \geq 0} (-t)^k j^{- k} $$ to obtain $$ \sum_{j \geq 1} [ (j + t)^{-1} - j^{-1} ] = \sum_{j \geq 1} \sum_{k \geq 1} (-t)^k j^{-k-1} = \sum_{k \geq 1} \zeta (k + 1) (-t)^k . $$