<s>[INST] Why is $\sum \frac{\sqrt{n^3+n}-\sqrt{n^3}}{\sqrt{n(n+1)}-1}$ convergent? We have the following series:
$$\sum_{n=1}^{\infty} \frac{\sqrt{n^3+n}-\sqrt{n^3}}{\sqrt{n(n+1)}-1}$$
According to WolframAlpha it is convergent but I don't see why. Intuitively, the expression under the sum must be assymptotically similair to something like $\frac{1}{n^\alpha}$, and clearly $\sqrt{n^2}$ dominates the denominator. In the numerator the expression looks similair either to $\sqrt{n^3}$ but $\frac{\sqrt{n^3}}{\sqrt{n^2}}={\sqrt{n}}$ which gives as assymptotic similarity to sequence, the sum of which is divergent. On the other hand, if $\sqrt{n}$ dominates the numerator  we get $\frac{\sqrt{n}}{\sqrt{n^2}}=\frac{1}{\sqrt{n}}$ so again assymptotic similarity to sequence, which gives divergent sum.
So why is this sum convergent anyway?
 [/INST] You may write, as $n$ is great:
$$
\begin{align}
\frac{\sqrt{n^3+n}-\sqrt{n^3}}{\sqrt{n(n+1)}-1}&=\frac{(\sqrt{n^3+n}-\sqrt{n^3})(\sqrt{n(n+1)}+1)}{n(n+1)-1}\\\\
&=\frac{n^{5/2} }{n^2}\frac{\left(\sqrt{1+\frac{1}{n^2}}-1\right)\left(\sqrt{1+\frac{1}{n^2}}+\frac1n\right)}{1+\frac{1}{n}-\frac{1}{n^2}}\\\\
&=\frac{n^{5/2} }{n^2}\frac{\left(\frac{1}{n^2}+\mathcal{O}\left(\frac{1}{n^3}\right)\right)\left(1+\mathcal{O}\left(\frac{1}{n}\right)\right)}{1+\frac{1}{n}-\frac{1}{n^2}}\\\\
& \sim \frac{n^{5/2}}{n^4}=\frac{1}{n^{3/2} }
\end{align}
$$ and your series is convergent by the comparison test.
</s>