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?
 A: Your evaluation is incorrect. For instance, consider the following. We have
$$\underbrace{\dfrac1n + \dfrac1n + \dfrac1n + \cdots + \dfrac1n}_{n \text{ times}} = 1$$
If we were to apply an argument similar to yours, since $\lim_{n \to \infty} \dfrac1n =0$, we have $$1 = \lim_{n \to \infty} 1 = \lim_{n \to \infty} \left(\dfrac1n + \cdots + \dfrac1n \right) = \lim_{n \to \infty} \dfrac1n + \cdots +\lim_{n \to \infty} \dfrac1n = 0 + \cdots + 0 = 0$$which is clearly false.
The fact that "the limit of the sum is sum of the limits" is only true for finite sums and not for infinite sums i.e.
$$\lim_{n \to \infty} \sum_{k=1}^{m} f_k(n) = \sum_{k=1}^m \lim_{n \to \infty} f_k(n)$$ is true only when $\lim_{n \to \infty} f_k(n)$ exists as a a real number and more importantly when '$m$' is a constant natural number independent of $n$.
Note that
$$ \sum_{k=1}^n \dfrac1{\sqrt{n^3+3n}} \leq \sum_{k=1}^n \dfrac1{\sqrt{n^3+3k}} \leq \sum_{k=1}^n \dfrac1{\sqrt{n^3+3}}$$
Hence, we get that
$$ \dfrac{n}{\sqrt{n^3+3n}} \leq \sum_{k=1}^n \dfrac1{\sqrt{n^3+3k}} \leq \dfrac{n}{\sqrt{n^3+3}} < \dfrac1{\sqrt{n}}$$
Now use squeeze theorem to obtain the limit.
A: 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 $\lim_{n\to\infty} T_n =0$. 
A: 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 $0$.
