Convergence of $sum limits_{n=1}^{infty}sqrt{n^3+1}-sqrt{n^3-1}$

Note that$$frac{1}{n^2} cdot frac{2sqrt{n}}{sqrt{1+frac{1}{n^3}}+sqrt{1-frac{1}{n^3}}}=frac{1}{n^{3/2}} cdotunderbrace{ frac{2}{sqrt{1+frac{1}{n^3}}+sqrt{1-frac{1}{n^3}}}}_{<frac{2}{1+0}=2}<2n^{-3/2}.$$

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 $ngeq1$,

$$sqrt{n^3+1}-sqrt{n^3-1}=left(sqrt{1+frac2{n^3-1}}-1right)sqrt{n^3-1}lefrac{sqrt{n^3-1}}{n^3-1}=(n^3-1)^{-frac12}le n^{-frac32},$$

where I have used the useful fact that $sqrt{1+x}lesqrt{1+x+frac{x^2}4}=1+frac x2$ for all $xgeq-1$.

You cannot prove $$frac{2sqrt{n}}{sqrt{1+frac{1}{n^3}}+sqrt{1-frac{1}{n^3}}} leq 1 $$ 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: $$ frac{2}{sqrt{n^3+1}+sqrt{n^3-1}}=frac{2}{n^{3/2}} cdot frac{1}{sqrt{1+frac{1}{n^3}}+sqrt{1-frac{1}{n^3}}} <frac{2}{n^{3/2}} , . $$