Answer
$\dfrac{1}{p+1}$
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
$S_n=\Sigma_{k=1}^{n} (\dfrac{1}{k+p}-\dfrac{1}{k+p+1})$
or, $= (\dfrac{1}{p+1}-\dfrac{1}{p+2})+ (\dfrac{1}{p+2}-\dfrac{1}{p+3})+.......+ (\dfrac{1}{n+p}-\dfrac{1}{n+p+1})$
or, $= \dfrac{1}{p+1}-\dfrac{1}{n+p+1}$
Now, $S=\lim\limits_{n \to \infty} S_n=\lim\limits_{n \to \infty} \dfrac{1}{p+1}-\lim\limits_{n \to \infty} \dfrac{1}{n+p+1}=\dfrac{1}{p+1}-0$
or, $S=\dfrac{1}{p+1}$
