Answer
$$
\sum_{n=1}^{\infty} \frac{1}{(2n+3)^{3}}
$$
Apply the Integral Test to the given series we obtain:
$$
\int_{1}^{\infty} \frac{1}{(2x+3)^{3}} d x=\frac{1}{100} .
$$
So, the series converges
Work Step by Step
$$
\sum_{n=1}^{\infty} \frac{1}{(2n+3)^{3}}
$$
Apply the Integral Test to the series
$$
\sum_{n=1}^{\infty} \frac{1}{(2n+3)^{3}}
$$
The function
$$
f(x)=\frac{1}{(2x+3)^{3}}
$$
is positive and continuous for $x \geq 1$ . To determine whether $f$ is decreasing, find the derivative.
$$
f^{'}(x)= \frac{-6}{(2x+3)^{4}}.
$$
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
$$
\begin{aligned}
\int_{1}^{\infty} \frac{1}{(2x+3)^{3}} d x &=\lim\limits_{t \to \infty } \int_{1}^{t } \frac{1}{(2x+3)^{3}} d x \\
&=\lim\limits_{t \to \infty } [\frac{-1}{4(2x+3)^{2}}]_{1}^{t} \\
&=0+[\frac{1}{4(2+3)^{2}}]\\
&=\frac{1}{100} .
\end{aligned}
$$
So, the series converges.