Answer
converges
Work Step by Step
Given $$\sum_{n=1}^{\infty}\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)$$
Since
$$\sum_{n=1}^{\infty}\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)=\left(1-\frac{1}{\sqrt{2}}\right)+\left(\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}\right)+\left(\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{4}}\right)+\ldots $$
Then the $n^{th}$ partial sum is given by
\begin{align*}
s_{n}&=\left(1-\frac{1}{\sqrt{2}}\right)+\left(\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}\right)+\left(\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{4}}\right)+\cdots+\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\\
&=1-\frac{1}{\sqrt{n+1}}
\end{align*}
Hence
\begin{align*}
S&=\lim _{n \rightarrow \infty} S_{n}\\
&=1-\lim _{n \rightarrow \infty} \frac{1}{\sqrt{n+1}}\\
&=1
\end{align*}
Thus, the series converges.
