Test the convergence of the following series: \frac{\sqrt{n+1}-1}{(n+2)^3-1}+...+\infty

jubateee 2022-01-07 Answered
Test the convergence of the following series:
\(\displaystyle{\frac{{\sqrt{{{n}+{1}}}-{1}}}{{{\left({n}+{2}\right)}^{{3}}-{1}}}}+\ldots+\infty\)

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Expert Answer

limacarp4
Answered 2022-01-08 Author has 2997 answers
Since
\(\displaystyle{\frac{{\sqrt{{{n}+{1}}}-{1}}}{{{\left({n}+{2}\right)}^{{3}}-{1}}}}={\frac{{{1}}}{{{n}^{{\frac{{5}}{{2}}}}}}}{\frac{{\sqrt{{{1}+\frac{{1}}{{n}}}}-\sqrt{{\frac{{1}}{{n}}}}}}{{{\left({1}+\frac{{2}}{{n}}\right)}^{{3}}-\frac{{1}}{{n}^{{3}}}}}}\)
\(\displaystyle\sim{\frac{{{1}}}{{{n}^{{\frac{{5}}{{2}}}}}}}\)
your method looks fine.
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Ethan Sanders
Answered 2022-01-09 Author has 918 answers
\(\displaystyle{\frac{{\sqrt{{{n}+{1}}}-{1}}}{{{\left({n}+{2}\right)}^{{3}}-{1}}}}={\frac{{{\left(\sqrt{{{n}+{1}}}-{1}\right)}{\left(\sqrt{{{n}+{1}}}\right\rbrace}+{1}}}{{{\left({\left({n}+{2}\right)}^{{3}}-{1}\right)}{\left(\sqrt{{{n}+{1}}}+{1}\right)}}}}\)
\(\displaystyle{\frac{{{n}}}{{{\left({\left({n}+{2}\right)}^{{3}}-{1}\right)}{\left(\sqrt{{{n}+{1}}}+{1}\right)}}}}{ < }{\frac{{{n}}}{{{n}^{{3}}}}}\)
\(\displaystyle={\frac{{{1}}}{{{n}^{{2}}}}}\)
Using the p-test, \(\displaystyle{\frac{{{1}}}{{{n}^{{2}}}}}\) converges, and since \(\displaystyle{0}{ < }{\frac{{{n}}}{{{\left({\left({n}+{2}\right)}^{{3}}-{1}\right)}{\left(\sqrt{{{n}+{1}}}+{1}\right)}}}}{ < }{\frac{{{1}}}{{{n}^{{2}}}}}\), using the comparison test, the original series must also converge.
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karton
Answered 2022-01-11 Author has 8659 answers

It is clear that \(\frac{\sqrt{n+1}-1}{(n+2)^3-1}\sim\frac{1}{n^{\frac{5}{2}}}\) as \(n\to\infty\). Since the series \(\sum_{n=1}^\infty\frac{1}{n^2}\) converges so that the series must be converge

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