TITLE: Series $\sum_{k=1}^\infty \left(\frac{1}{k}-\frac{1}{k+z} \right)$
QUESTION [11 upvotes]: If $z$ is an integer, the sum of the series
  $$\sum_{k=1}^\infty \left(\frac{1}{k}-\frac{1}{k+z}\right)$$ is easy since it is a telescoping series. But if $z$ is a fraction, say $z=3/2$, I don't see why the series sums to $$\frac{8}{3}-\ln 4$$
Is there a formula for $z=m/n$, where $m,n$ are positive integers and $n\neq 1$?

REPLY [3 votes]: Simple.
$$\begin{align}\sum_{k=1}^\infty\frac1k-\frac1{k+\frac32}&=-2\sum_{k=1}^\infty\frac1{2k+3}-\frac1{2k}\\&=\frac83-2\sum_{k=1}^\infty\frac1{2k-1}-\frac1{2k}\\&=\frac83+2\sum_{k=1}^\infty\frac{(-1)^{k+1}}k\\&=\frac83-2\ln(2)\\&=\frac83-\ln(4)\end{align}$$
This approach works for any $z=\frac12+n,~n\in\Bbb N$.