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

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

• Live experts 24/7
• Questions are typically answered in as fast as 30 minutes
• Personalized clear answers

### Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

limacarp4
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.
###### Not exactly what youâ€™re looking for?
Ethan Sanders
$$\displaystyle{\frac{{\sqrt{{{n}+{1}}}-{1}}}{{{\left({n}+{2}\right)}^{{3}}-{1}}}}={\frac{{{\left(\sqrt{{{n}+{1}}}-{1}\right)}{\left(\sqrt{{{n}+{1}}}\right\rbrace}+{1}}}{{{\left({\left({n}+{2}\right)}^{{3}}-{1}\right)}{\left(\sqrt{{{n}+{1}}}+{1}\right)}}}}$$
$$\displaystyle{\frac{{{n}}}{{{\left({\left({n}+{2}\right)}^{{3}}-{1}\right)}{\left(\sqrt{{{n}+{1}}}+{1}\right)}}}}{ < }{\frac{{{n}}}{{{n}^{{3}}}}}$$
$$\displaystyle={\frac{{{1}}}{{{n}^{{2}}}}}$$
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.
karton

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