<s>[INST] Does the series $\sum_{n=1}^{\infty} (n^2+1)^{1/2}-(n^3+1)^{1/3}$ converge? Does the series $\sum_{n=1}^{\infty} (n^2+1)^{1/2}-(n^3+1)^{1/3}$ converge?
I am not able to use the standard tests on this series. Any hints/advice will be much appreciated. 
The solution hint is: 

I don't understand how one get's this expression. The first fraction is clear but the last fraction is not so clear. What did the author use to rationalize the expression? 
 [/INST] Since 'the first fraction is clear', the question about the hint seems to be if the equality below is true,
$$ n^2+1 - \sqrt[3]{(n^3+1)^2}=\frac{3n^4 - 2n^3+3n^2}{(n^2+1)^2+(n^2+1)\sqrt[3]{(n^3+1)^2} + (n^3+1)\sqrt{n^3+1}}$$
Observe that if we define $a=\sqrt{n^2+1}>0\, ,b=\sqrt[3]{n^3+1}>0$, this is asking if
$$ a^2 - b^2 = \frac{3n^4 - 2n^3 + 3n^2}{a^4 + a^2 b^2 + b^4} $$
which is equivalent to
$$ (a^2-b^2)(a^4 + a^2b^2 + b^4) =3n^4 - 2n^3 + 3n^2$$
we rewrite the left hand side,
\begin{align}
LHS
&= (a^2-b^2)(a^4 + a^2b^2 + b^4)  \\&= a^6 + \color{red}{a^4b^2}+\color{blue}{a^2b^4}-\color{red}{a^4b^2}-\color{blue}{a^2b^4}-b^6\\
&= a^6 - b^6\\
&=(n^2+1)^3 - (n^3+1)^2\\
&= (n^6+3n^4+3n^2+1) - (n^6 + 2n^3 + 1)\\
&= 3n^4 - 2n^3 + 3n^2\\
&= RHS
\end{align}
as needed. I much prefer the other answers, but there you go.
</s>