# 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}

Question
Series

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.
$$\displaystyle{\sum_{{{n}={1}}}^{\infty}}{\frac{{{1}}}{{{\left({2}{n}+{3}\right)}^{{3}}}}}$$

2020-10-19
Integral test:
Let f(x) is positive decreasing continuous function on $$\displaystyle{\left[{k},\infty\right)}$$ and $$\displaystyle{f{{\left({n}\right)}}}={a}_{{n}}$$ then
If $$\displaystyle{\int_{{k}}^{\infty}}{f{{\left({x}\right)}}}{\left.{d}{x}\right.}$$ is convergent then $$\displaystyle{\sum_{{{n}={k}}}^{\infty}}{a}_{{n}}$$ is also convergent.
If $$\displaystyle{\int_{{k}}^{\infty}}{f{{\left({x}\right)}}}{\left.{d}{x}\right.}$$ is divergent then $$\displaystyle{\sum_{{{n}={k}}}^{\infty}}{a}_{{n}}$$ is also divergent.
So we can use integral test in given function.
Given that
$$\displaystyle{\sum_{{{n}={1}}}^{\infty}}{\frac{{{1}}}{{{\left({2}{n}+{3}\right)}^{{3}}}}}$$
Integral test:
Here $$\displaystyle{f{{\left({x}\right)}}}={\frac{{{1}}}{{{\left({2}{n}+{3}\right)}^{{3}}}}}$$ so
$$\displaystyle{\int_{{{1}}}^{\infty}}{\frac{{{1}}}{{{\left({2}{n}+{3}\right)}^{{3}}}}}{\left.{d}{x}\right.}={\int_{{{1}}}^{\infty}}{\left({2}{x}+{3}\right)}^{{-{3}}}{\left.{d}{x}\right.}={\frac{{{1}}}{{{2}}}}{{\left[{\frac{{{\left({2}{x}+{3}\right)}^{{-{2}}}}}{{-{2}}}}\right]}_{{1}}^{\infty}}$$
$$\displaystyle=-{\frac{{{1}}}{{{4}}}}{{\left[{\frac{{{1}}}{{{\left({2}{x}+{3}\right)}^{{2}}}}}\right]}_{{1}}^{\infty}}$$
$$\displaystyle=-{\frac{{{1}}}{{{4}}}}{\left[{\frac{{{1}}}{{{25}}}}-{0}\right]}$$
$$\displaystyle={\frac{{{1}}}{{{100}}}}$$
Which is finite, so $$\displaystyle{\int_{{1}}^{\infty}}{\frac{{{1}}}{{{\left({2}{x}+{3}\right)}^{{3}}}}}{\left.{d}{x}\right.}$$ is convergent
Since $$\displaystyle{\int_{{1}}^{\infty}}{\frac{{{1}}}{{{\left({2}{x}+{3}\right)}^{{3}}}}}{\left.{d}{x}\right.}$$ is convergent so using integral test $$\displaystyle{\sum_{{{n}={1}}}^{\infty}}{\frac{{{1}}}{{{\left({2}{n}+{3}\right)}^{{3}}}}}$$ si also convergent.

### Relevant Questions

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