What is the formula to find $a_k = \frac{1}{k} + \frac{1}{2(k+1)}+ \frac{1}{3(k+2)} + \dots $ for any $k \in \mathbb{N}^+$?

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*}

Hint. For $k > 1$ write the general term using partial fractions and look for eventual telescoping.