<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] The answer is correct, the reasoning is not.  First we solve the problem. 
The sum has $n$ terms. Each term is $\lt \frac{1}{n^{3/2}}$. It follows that if $S_n$ is our sum, then
$$0\lt S_n \lt \frac{n}{n^{3/2}}=\frac{1}{n^{1/2}}.$$
Now let $n\to\infty$. Note that $\frac{1}{n^{1/2}}\to 0$, so by Squeezing $S_n\to 0$.
Remark: It is true that each term in the sum $S_n$ approaches $0$. However, the number of terms in the sum increases. Note for example that the sum
$$T_n=\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2n}$$
does not approach $0$, for clearly the sum is always $\ge \frac{1}{2}$. Yet if we applied the "$0+0+\cdots+0$" reasoning, we would conclude incorrectly that $\lim_{n\to\infty} T_n =0$. 
</s>