What is the formula to find $a_k = \frac{1}{k} + \frac{1}{2(k+1)}+ \frac{1}{3(k+2)} + \dots $ for any $k \in \mathbb{N}^+$? ...in other words, how can I write
\begin{equation}
a_k = \sum_{n=k}^\infty \frac{1}{n(n-k+1)}
\end{equation}
in a simpelr way? If $k=1$ this is the hard but well know Basel problem: Euler says $a_1=\frac{\pi^2}{6}$. If $k=2$ we have the Mengoli serie (usually written as $\sum_{n=1}^\infty = \frac{1}{n(n+1)}$, but $\sum_{n=2}^\infty = \frac{1}{n(n-1)}$ is the same), and it is easy to see that $a_2=1$. But I can't generalize the result. Other values of the sequence are $a_3=\frac{3}{4}$, $a_4=\frac{11}{18}$, $a_5=\frac{25}{48}$, $a_6=\frac{137}{300}$, $a_7=\frac{49}{120}$, $a_8=\frac{363}{980}$ (I found these with WolframAlpha, but it can't find the general term). I can't guess a formula to be checked by induction. I can't see a pattern and after all looking to these fraction it is clear that some simplifications occur so the online encyclopedia of integer sequences is useless.
Edit
I wrote the serie in this way but there aren't telescopic simplifications. I'm stuck

 A: We have
\begin{align*}
\sum_{n=k}^\infty\frac1{n(n-k+1)}&=\sum_{n=0}^\infty\frac1{(n+k)(n+1)}\\
&=\frac{1}{k-1}\sum_{n=0}^\infty\left[\frac1{n+1}-\frac1{n+k}\right]\\
&=\frac{1}{k-1}\lim_{m\to\infty}\sum_{n=0}^{k+m}\left[\frac1{n+1}-\frac1{n+k}\right]\\
\end{align*}
Notice that
\begin{align*}
\sum_{n=0}^{k+m}\frac1{n+1}&=1+\frac{1}{2}+\dots+\frac{1}{k-1}+\color{red}{\frac1{k}+\frac1{k+1}+\dots+\frac1{k+m+1}}\\
\sum_{n=0}^{k+m}\frac1{n+k}&=\color{red}{\frac1k+\frac1{k+1}+\dots+\frac1{k+m+1}}+\frac1{k+m+2}+\dots+\frac1{2k+m-1}+\frac1{2k+m}
\end{align*}
Therefore,
\begin{align*}
\sum_{n=0}^{k+m}\left[\frac1{n+1}-\frac1{n+k}\right]&=1+\frac 12+\dots+\frac1{k-1}-\frac1{k+m+2}-\dots-\frac1{2k+m}\\
&=H_{k-1}-\sum_{n=2}^k\frac1{k+m+n},
\end{align*}
where $H_{k-1}$ is the $(k-1)$th harmonic number. Hence,
\begin{align*}
\frac{1}{k-1}\lim_{m\to\infty}\sum_{n=0}^{k+m}\left[\frac1{n+1}-\frac1{n+k}\right]&=\frac{1}{k-1}\lim_{m\to\infty}\left[H_{k-1}-\sum_{n=2}^k\frac1{k+m+n}\right]\\
&=\frac{H_{k-1}}{k-1}.
\end{align*}
A: Hint. For $k > 1$ write the general term using partial fractions and look for eventual telescoping.
