Confirm that the Integral Test can be applied to the series. Then use the Integral Test to determine the convergence or divergence of the series.sum_{n=1}^inftyfrac{1}{(2n+3)^3}

Integral test:
Let f(x) is positive decreasing continuous function on ${\displaystyle [k,\mathrm{\infty })}$ and ${\displaystyle f\left(n\right)=a_n}$ then
If ${\displaystyle {\int }_k^{\mathrm{\infty }}f\left(x\right)dx}$ is convergent then ${\displaystyle \sum _{n=k}^{\mathrm{\infty }}{a}_n}$ is also convergent.
If ${\displaystyle {\int }_k^{\mathrm{\infty }}f\left(x\right)dx}$ is divergent then ${\displaystyle \sum _{n=k}^{\mathrm{\infty }}{a}_n}$ is also divergent.
So we can use integral test in given function.
Given that
${\displaystyle \sum _{n=1}^{\mathrm{\infty }}\frac{1}{{(2n+3)}^3}}$
Integral test:
Here ${\displaystyle f\left(x\right)=\frac{1}{{(2n+3)}^3}}$ so
${\displaystyle {\int }_1^{\mathrm{\infty }}\frac{1}{{(2n+3)}^3}dx={\int }_1^{\mathrm{\infty }}{(2x+3)}^{-3}dx=\frac{1}{2}{\left[\frac{{(2x+3)}^{-2}}{-2}\right]}_1^{\mathrm{\infty }}}$
${\displaystyle =-\frac{1}{4}{\left[\frac{1}{{(2x+3)}^2}\right]}_1^{\mathrm{\infty }}}$
${\displaystyle =-\frac{1}{4}[\frac{1}{25}-0]}$
${\displaystyle =\frac{1}{100}}$
Which is finite, so ${\displaystyle {\int }_1^{\mathrm{\infty }}\frac{1}{{(2x+3)}^3}dx}$ is convergent
Since ${\displaystyle {\int }_1^{\mathrm{\infty }}\frac{1}{{(2x+3)}^3}dx}$ is convergent so using integral test ${\displaystyle \sum _{n=1}^{\mathrm{\infty }}\frac{1}{{(2n+3)}^3}}$ si also convergent.