Question

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

Answer 1

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.