jubateee

2022-01-07

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

limacarp4

Expert

Since
$\frac{\sqrt{n+1}-1}{{\left(n+2\right)}^{3}-1}=\frac{1}{{n}^{\frac{5}{2}}}\frac{\sqrt{1+\frac{1}{n}}-\sqrt{\frac{1}{n}}}{{\left(1+\frac{2}{n}\right)}^{3}-\frac{1}{{n}^{3}}}$
$\sim \frac{1}{{n}^{\frac{5}{2}}}$
your method looks fine.

Ethan Sanders

Expert

$\frac{\sqrt{n+1}-1}{{\left(n+2\right)}^{3}-1}=\frac{\left(\sqrt{n+1}-1\right)\left(\sqrt{n+1}\right\}+1}{\left({\left(n+2\right)}^{3}-1\right)\left(\sqrt{n+1}+1\right)}$
$\frac{n}{\left({\left(n+2\right)}^{3}-1\right)\left(\sqrt{n+1}+1\right)}<\frac{n}{{n}^{3}}$
$=\frac{1}{{n}^{2}}$
Using the p-test, $\frac{1}{{n}^{2}}$ converges, and since $0<\frac{n}{\left({\left(n+2\right)}^{3}-1\right)\left(\sqrt{n+1}+1\right)}<\frac{1}{{n}^{2}}$, using the comparison test, the original series must also converge.

karton

Expert

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