<s>[INST] Is my limit evaluation correct? I'm studying for my calculus exam and I have the following limit:
$$\lim\limits_{n \to \infty} \left ( \frac{1}{\sqrt{n^3 +3}}+\frac{1}{\sqrt{n^3 +6}}+ \cdots +\frac{1}{\sqrt{n^3 +3n}} \right )$$
My solution is:
$$\begin{align*}
&\lim\limits_{n\ \to \infty} \left ( \frac{1}{\sqrt{n^3 +3}}+\frac{1}{\sqrt{n^3 +6}}+ \cdots +\frac{1}{\sqrt{n^3 +3n}} \right )\\&= \lim_{n \to \infty}\left ( \frac{1}{\sqrt{n^3 +3}} \right ) + \lim_{n \to \infty}\left ( \frac{1}{\sqrt{n^3 + 6}} \right ) + \cdots + \lim_{n \to \infty} \left ( \frac{1}{\sqrt{n^3 +3n}} \right )\\
&=0 + 0 + \cdots + 0\\
&= 0
\end{align*}$$
It turned out to be suspiciously easy to solve.
Is this correct? If it isn't, what is wrong and how can I solve it correctly?
 [/INST] As $n$ grows, the number of terms increases. So you need extra care.
The sum is equal to $\sum_{k=1}^n \frac{1}{\sqrt{n^3+3k}}$.
We have 
$\left | \sum_{k=1}^n \frac{1}{\sqrt{n^3+3k}}\right| \leq n \cdot \frac{1}{\sqrt{n^3}}=\frac{1}{\sqrt{n}}$.
Now you can take a limit and the limit is $0$.
</s>