<s>[INST] Series $\sum_{k=1}^\infty \left(\frac{1}{k}-\frac{1}{k+z} \right)$ 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 $n\neq 1$?
 [/INST] For $\vert x \vert < 1$, if we define $f(x)$ as$$f(x)=\sum_{n=0}^{\infty} \left(x^n - x^{n+z}\right)$$ then $$\int_0^1 f(x) dx = \sum_{n=1}^{\infty} \left(\dfrac1n - \dfrac1{n+z}\right)$$
But we have $$f(x) = (1-x^z) \sum_{n=0}^{\infty} x^n = \dfrac{1-x^z}{1-x}$$
Hence,
$$\sum_{n=1}^{\infty} \left(\dfrac1n - \dfrac1{n+z}\right) = \int_0^1 \dfrac{1-x^z}{1-x}dx = I(z)$$
For $z=3/2$, we have
\begin{align}
I(3/2) & = \overbrace{\int_0^1 \dfrac{1-x^{3/2}}{1-x}dx = \int_0^1 \dfrac{1-t^3}{1-t^2} \cdot 2tdt}^{\text{Substitute }x = t^2} = \int_0^1 \dfrac{2t(1+t+t^2)}{1+t}dt\\
& = \int_0^1 \dfrac{2t((1+t)^2-t)}{1+t}dt = \int_0^1 2t(1+t) dt - 2 \int_0^1 \dfrac{t^2}{1+t} dt\\
& = \left(t^2 + \dfrac{2t^3}3\right)_{t=0}^{t=1} - 2 \int_0^1 \dfrac{t^2-1}{t+1}dt - 2 \int_0^1 \dfrac{dt}{1+t}\\
& = \dfrac53 - 2 \left(\dfrac{t^2}2 - t\right)_{t=0}^{t=1}-2\ln(2)\\
& = \dfrac53 - 2 \left(\dfrac12-1\right) - \ln(4) = \dfrac83 - \ln(4)
\end{align}
</s>