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
 A: Hint $:$ Observe that $\frac {1} {\sqrt {n^3+1}} < \frac {1} {n^{\frac 3 2}}$ for all $n \geq 1$ and $\sum\limits_{n=1}^{\infty} \frac {1} {n^{\frac 3 2}} < \infty.$
A: Use the fact that$$\lim_{n\to\infty}\frac{\dfrac1{\sqrt{n^3+1}}}{\dfrac1{n^{3/2}}}=1.$$
A: A polynomial is asymptotically equivalent to its leading term, so $n^3+1\sim_\infty n^3$, whence
$$\frac1{\sqrt{n^3+1}}\sim_\infty\frac1{n^{3/2}},$$
and the latter is convergent.
A: By the limit comparison test, you should have that $$a_n=\frac{1}{\sqrt{n^3+1}}\quad\text{and}\quad  b_n=\frac{1}{\sqrt{n^3}},$$
and now we take the limit of $a_n\div b_n$, as thusly demonstrated:
$$\begin{align}\lim_{n\to\infty}\frac{a_n}{b_n}&=\lim_{n\to\infty}\frac{\frac{1}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n^3}}} \\ &= \lim_{n\to\infty}\frac{\sqrt{n^3}}{\sqrt{n^3+1}} \\ &=\lim_{n\to\infty}\frac{1}{\sqrt{1+\frac{1}{n^3}}}.\end{align}$$
Since $\lim\limits_{n\to\infty}\frac{1}{n^3}=0$, it follows that  $$\lim_{n\to\infty}\frac{a_n}{b_n}=\frac{1}{\sqrt{1+0}}=\frac{1}{1+0} = 1 > 0$$ and by that, we can conclude that this limit exists, for which $\sum\limits_{n=1}^\infty\dfrac{1}{\sqrt{n^3}}$ converges.
$$\therefore \sum_{n=1}^\infty\frac{1}{\sqrt{n^3+1}}\;\text{also}\; \boxed{\text{converges!}}$$
