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? 
 A: I take this chance as a teaching opportunity.
Convergence tests are just condensed forms of chains of inequalities. As a thumb rule, if you cannot figure out which convergence test is best suited for the task at hand, it is worth to take a step back and just try to tackle your problem by elementary inequalities. First consideration: for any $n\geq 1$, both $\sqrt{n^2+1}$ and $\sqrt[3]{n^3+1}$ are pretty close to $n$, so it looks reasonable that the convergence of the given series depends on how much they are close to $n$. We have
$$\sqrt{n^2+1}-n = \frac{1}{n+\sqrt{n^2+1}}\approx \frac{1}{2n} $$
$$\sqrt[3]{n^3+1}-n = \frac{1}{n^2+n\sqrt[3]{n^3+1}+\sqrt[3]{n^3+1}^2}\approx \frac{1}{3n^2} $$
due to $(a^2-b^2)=(a-b)(a+b)$ and $(a^3-b^3)=(a-b)(a^2+ab+b^2)$. The series $\sum_{n\geq 1}\frac{1}{3n^2}$ is convergent, but the series $\sum_{n\geq 1}\frac{1}{n}$ is not, so the given series is expected to be divergent. In order to turn this second consideration into a rigorous proof, it is enough to replace the first approximation above with
$$\forall n\geq 1,\qquad \sqrt{n^2+1}-n \geq \frac{1}{3n} $$
implying that $\sum_{n\geq 1}\sqrt{n^2+1}-\sqrt[3]{n^3+1}$ is really divergent.
A: Use development of in Taylor Series, you will have your result.
For example
$$
\sqrt[3]{1+n^3}=n\sqrt[3]{1+\frac{1}{n^3}}=n\left(1+\frac{1}{3n^3}+o\left(\frac{1}{n^3}\right)\right)
$$
Do the same for the square root and tell us if it converges or not.
Identically,
$$
\sqrt{1+n^2}=n\sqrt[2]{1+\frac{1}{n^2}}=n\left(1+\frac{1}{2n^2}+o\left(\frac{1}{n^2}\right)\right)
$$
Hence
$$
\sqrt{1+n^2}-\sqrt[3]{1+n^3}\underset{(+\infty)}{=}n+\frac{1}{2n}-n-\frac{1}{3n^2}+o\left(\frac{1}{n}\right)
$$
Hence sadly

$$
\sqrt{1+n^2}-\sqrt[3]{1+n^3}\underset{(+\infty)}{\sim}\frac{1}{2n}$$

The series $\displaystyle \sum_{ \geq 1}^{ }\frac{1}{n}$ diverges so as your series.
A: 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.
