Convergence of $\sum \limits_{n=1}^{\infty}\sqrt{n^3+1}-\sqrt{n^3-1}$ Hello I am a high school student from germany and I am starting to study math this october. I am trying to prepare myself for the analysis class which I will attend so I got some analysis problems from my older cousin who also studied maths. But I am stuck on this problem.
Check the following series for convergence/divergence $$\sum \limits_{n=1}^{\infty}\sqrt{n^3+1}-\sqrt{n^3-1}$$ I tried to prove the convergence by comparison test  $$\sqrt{n^3+1}-\sqrt{n^3-1}= \frac{2}{\sqrt{n^3+1}+\sqrt{n^3-1}}=\frac{1}{n^2} \cdot \frac{2\sqrt{n}}{\sqrt{1+\frac{1}{n^3}}+\sqrt{1-\frac{1}{n^3}}}$$ and then compare it with $$\sum \limits_{n=1}^{\infty}\frac{1}{n^2}$$  But in order to do that, I need to prove that $$\frac{2\sqrt{n}}{\sqrt{1+\frac{1}{n^3}}+\sqrt{1-\frac{1}{n^3}}} \leq 1$$ But I am having problems to prove that. Does anyone have tip how to solve this problem?
 A: You cannot prove
$$\frac{2\sqrt{n}}{\sqrt{1+\frac{1}{n^3}}+\sqrt{1-\frac{1}{n^3}}} \leq 1
$$
because the left-hand side tends to $+\infty$ for $n \to \infty$.
You should compare your series with $\sum_{n=1}^{\infty}\frac{1}{n^{3/2}}$ instead:
$$
\frac{2}{\sqrt{n^3+1}+\sqrt{n^3-1}}=\frac{2}{n^{3/2}} \cdot \frac{1}{\sqrt{1+\frac{1}{n^3}}+\sqrt{1-\frac{1}{n^3}}} <\frac{2}{n^{3/2}} \, .
$$
A: As Martin R said, you might want to compare it to $\sum_{n=1}^\infty n^{-\frac32}<\infty$. Your rationalization approach is completely fine for this. One could also say that for $n\geq1$,
$$\sqrt{n^3+1}-\sqrt{n^3-1}=\left(\sqrt{1+\frac2{n^3-1}}-1\right)\sqrt{n^3-1}\le\frac{\sqrt{n^3-1}}{n^3-1}=(n^3-1)^{-\frac12}\le n^{-\frac32},$$
where I have used the useful fact that $\sqrt{1+x}\le\sqrt{1+x+\frac{x^2}4}=1+\frac x2$ for all $x\geq-1$.
A: Note that$$\frac{1}{n^2} \cdot \frac{2\sqrt{n}}{\sqrt{1+\frac{1}{n^3}}+\sqrt{1-\frac{1}{n^3}}}=\frac{1}{n^{3/2}} \cdot\underbrace{ \frac{2}{\sqrt{1+\frac{1}{n^3}}+\sqrt{1-\frac{1}{n^3}}}}_{<\frac{2}{1+0}=2}<2n^{-3/2}.$$
