Answer
$\dfrac{1}{a(a+1)}$
Work Step by Step
Let us consider an infinite series $\Sigma_{n=k}^{n=\infty} a_n$. This series will converge to the sum $S$ when the sequence of its partial sums $\{S_n\}$ converge to $S$.
That is, $S=\lim\limits_{N \to \infty} S_n$ and $S=\Sigma_{n=k}^{n=\infty} a_n$.
This series will diverge when the limit does not exist and the series will diverge to infinity when the limit is infinite.
Here, we have
$S_n=\dfrac{1}{a} \Sigma_{k=1}^{n} (\dfrac{1}{ak+1}-\dfrac{1}{ak+a+1})$
or, $= \dfrac{1}{a} [\dfrac{1}{a+1}-\dfrac{1}{2a+1}+\dfrac{1}{2a+1}-\dfrac{1}{3a+1}+....+\dfrac{1}{na+1}-\dfrac{1}{(n+1)(a+1)}]$
or, $= \dfrac{1}{a} [\dfrac{1}{a+1}-\dfrac{1}{(n+1)(a+1)}]$
Now, $S=\lim\limits_{n \to \infty} S_n=\dfrac{1}{a} [\lim\limits_{n \to \infty} \dfrac{1}{a+1}-\lim\limits_{n \to \infty} \dfrac{1}{(n+1)(a+1)}=\dfrac{1}{a}[\dfrac{1}{a+1}-0]$
or, $S=\dfrac{1}{a(a+1)}$
