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