Answer
$$
\sum_{n=1}^{\infty} \frac{1}{n^{\frac{1}{2}}}
$$
Apply the Integral Test to the given series we obtain
$$
\int_{1}^{\infty} \frac{1}{x^{\frac{1}{2}}} d x =\infty.
$$
So, the series diverges
Work Step by Step
$$
\sum_{n=1}^{\infty} \frac{1}{n^{\frac{1}{2}}}
$$
Apply the Integral Test to the series
$$
\sum_{n=1}^{\infty} \frac{1}{n^{\frac{1}{2}}}
$$
The function
$$
f(x)=\frac{1}{x^{\frac{1}{2}}}
$$
is positive and continuous for $x\gt 1$ . To determine whether $f$ is decreasing, find the derivative.
$$
f^{'}(x)=\frac{-1}{2x^{\frac{3}{2}}}
$$
So, $f^{'}(x) \lt 0 $ for $x \gt k$,
and it follows that $f$ satisfies the conditions for the Integral Test.
You can integrate to obtain
$$
\begin{aligned}
\int_{1}^{\infty} \frac{1}{x^{\frac{1}{2}}} d x &=\lim\limits_{t \to \infty } \int_{1}^{t } \frac{1}{x^{\frac{1}{2}}} d x \\
&=\lim\limits_{t \to \infty } [2x^{\frac{1}{2}} ]_{1}^{t} \\
&=[2x^{\frac{1}{2}} ]_{1}^{\infty} \\
&=\infty
\end{aligned}
$$
So, the series diverges