 # 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} Cabiolab 2020-10-18 Answered

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.
$\sum _{n=1}^{\mathrm{\infty }}\frac{1}{{\left(2n+3\right)}^{3}}$

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