Answer
The infinite series diverges to infinity.
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
$S_n=\Sigma_{k=1}^{\infty} (\sqrt {k+1}-\sqrt k)$
or, $=(\sqrt {2}-\sqrt 1)+(\sqrt {3}-\sqrt 2)+(\sqrt {4}-\sqrt 3)+........(\sqrt {n}-\sqrt {n-1})+(\sqrt {n+1}-\sqrt n)$
or, $=\sqrt {n+1}-1$
Now, $S=\lim\limits_{n \to \infty} S_n=\lim\limits_{n \to \infty} \sqrt {n+1}-1=\infty$
This implies that the infinite series diverges to infinity.
