Series $\sum_{n=1}^{\infty} \frac{1}{\sqrt {n^3 + 1}}$ Alright, so I have a question regarding a series that I am attempting to solve:
$$\sum_{n=1}^{\infty} \frac{1}{\sqrt{n^3 + 1}}$$
To Start Off I attempted to use Direct Comparison Theorem:
$$\frac{1}{\sqrt {n^3}} < \frac{1}{\sqrt{n^3 + 1}}$$
Then, It's clear that this series converges by p series:
$$\sum_{n=1}^{\infty} \frac{1}{\sqrt {n^3}} = \sum_{n=1}^{\infty} \frac{1}{n^{3/2}}$$ 
But becuase this series is smaller, nothing can be assumed.
So then I moved onto Limit Comparison Theorem using:
$$b_n = \frac{1}{n^{3/2}}$$ 
But I got nowhere attempting to solve it:
$$\lim_{n->{\infty}} \frac{1}{\sqrt{n^3 + 1}}*{\frac{n^{3/2}}{1}} = \lim_{n->{\infty}} \frac{\sqrt {n^3}}{\sqrt{n^3 + 1}}$$
Which, once using L'Hospital's rule kinda just alternates I guess.
So my question: How do I solve this/where did I make the mistake?
Note: Only learned divergence tests up to Limit Comparison Theorem.
Only attempting to find if series is convergent or divergent

Hint $:$ Observe that $\frac {1} {\sqrt {n^3+1}} < \frac {1} {n^{\frac 3 2}}$ for all $n \geq 1$ and $\sum\limits_{n=1}^{\infty} \frac {1} {n^{\frac 3 2}} < \infty.$
