babyagelesszj

Answered

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 $

$\frac{\sqrt{2}-1}{{3}^{3}-1}+\frac{\sqrt{3}-1}{{4}^{3}-1}+\frac{\sqrt{4}-1}{{5}^{3}-1}+\cdots $

Answer & Explanation

Ordettyreomqu

Expert

2022-07-01Added 22 answers

Here, $\frac{\sqrt{n+1}-1}{(n+2{)}^{3}-1}\le \frac{1}{{n}^{2}}$. 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 $\sum _{1}^{\mathrm{\infty}}\frac{\sqrt{n+1}-1}{(n+2{)}^{3}-1}\le (\sum _{1}^{\mathrm{\infty}}\frac{1}{{n}^{2}}={\pi}^{2}/6)$. Thus the series converges.

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