babyagelesszj

2022-06-30

Want to test the series
$\frac{\sqrt{2}-1}{{3}^{3}-1}+\frac{\sqrt{3}-1}{{4}^{3}-1}+\frac{\sqrt{4}-1}{{5}^{3}-1}+\cdots$

Ordettyreomqu

Expert

Here, $\frac{\sqrt{n+1}-1}{\left(n+2{\right)}^{3}-1}\le \frac{1}{{n}^{2}}$. For n=1, it is true, and then the denominator of given series increases more rapidly than numerator and this rate is greater than the rate in $1/{n}^{2}$, so the given series is bounded by $1/{n}^{2}$ and hence $\sum _{1}^{\mathrm{\infty }}\frac{\sqrt{n+1}-1}{\left(n+2{\right)}^{3}-1}\le \left(\sum _{1}^{\mathrm{\infty }}\frac{1}{{n}^{2}}={\pi }^{2}/6\right)$. Thus the series converges.