<s>[INST] Want to test the series for convergence/divergence $\sum_n \frac{\sqrt{n+1} - 1}{(n+2)^3 - 1} $ Want to test the series $$\frac{\sqrt{2} - 1}{3^3 - 1} + \frac{\sqrt{3} - 1}{4^3 - 1} + \frac{\sqrt{4} - 1}{5^3 - 1} + \cdots$$ for convergence/divergence.
My attempt is
$U_n =\frac{\sqrt{n+1} - 1}{(n+2)^3 - 1} $ Now I wish to find $V_n $ which is less than (or greater than $U_n $) such that its convergence/divergence is known  so that I can find the behaviour of $U_n $
 [/INST] Here, $\frac{\sqrt{n+1}-1}{(n+2)^3-1} \leq \frac{1}{n^2}$. 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 $\sum_{1}^{\infty}\frac{\sqrt{n+1}-1}{(n+2)^3-1} \leq (\sum_{1}^{\infty}\frac{1}{n^2}=\pi^2/6)$. Thus the series converges. 
</s>