Answer
diverges
Work Step by Step
Since
\begin{align*}
\lim_{n\to \infty} a_n &=\lim_{n\to \infty } 1-\sqrt{1-\frac{1}{n}} \\
&=0
\end{align*}
and
\begin{align*}
a_n &= 1-\sqrt{1-\frac{1}{n}} \\
&=1-\sqrt{\frac{n-1}{n}}\\
&=\frac{\sqrt{n}-\sqrt{n-1}}{\sqrt{n}}\\
&=\frac{n-(n-1)}{\sqrt{n}(\sqrt{n}+\sqrt{n-1})}\\
&=\frac{1}{n+\sqrt{n^{2}-n}} \\
& \geq \frac{1}{n+\sqrt{n^{2}}}\\
&\geq \frac{1}{2 n}
\end{align*}
Since $\sum \frac{1}{2 n}$ diverges, then by using the comparison test, $\sum a_n $ also diverges.
