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? 

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.
