# Determine con-/divergence of $\sum\limits_{n=1}^\infty \left[ \sqrt{n^3+1} - n^{\frac32} \right]$ by (limit) comparison test

Problem:

Use the comparison test, or limit comparison test to determine whether the sum $\sum\limits_{n=1}^\infty \left[ \sqrt{n^3+1} - n^{\frac32} \right]$ converges or diverges.

My attempt:

Given my <1 day experience with this, it's hard for me to recognize candidates for comparison tests. A lot of the material I've read on this, takes for granted that we already know these "trivial" convergent series, making it look easy, but I don't have this luxury.

Don't get me wrong, I understand the comparison test, and the limit comparison test. But my repertoire of go-to comparisons is pretty shallow.

Any help appreciated!

$$\sqrt{n^3 + 1} - n^{3/2} = \frac{(n^3 + 1) - n^3}{\sqrt{n^3 + 1} + n^{3/2}} = \frac{1}{\sqrt{n^3 + 1} + n^{3/2}}.$$
$$\lim_{n\to \infty} \frac{n^{3/2}}{\sqrt{n^3 + 1} + n^{3/2}} = \lim_{n\to \infty} \frac{1}{\sqrt{1 + \frac{1}{n^3}} + 1} = \frac{1}{1 + 1} = \frac{1}{2}.$$
and $\sum_{n = 1}^\infty \frac{1}{n^{3/2}}$ converges (by the $p$-test with $p = 3/2$), by limit comparison, your series converges.