When is the limit of a sum equal to the sum of limits? I was trying to solve a problem and got stuck at the following step:
Suppose ${n \to \infty}$ .
$$\lim \limits_{n \to \infty} \frac{n^3}{n^3} = 1$$
Let us rewrite $n^3=n \cdot n^2$ as $n^2 + n^2 + n^2 + n^2 \dots +n^2$,$\space$ n times.
Now we have
$$\lim \limits_{n \to \infty} \frac{n^3}{n^3} = \frac {n^2 + n^2 + n^2 + n^2 + n^2 \dots +n^2}{n^3} $$
As far as I understand, we can always rewrite the limit of a sum as the sum of limits ...
$$\dots = \lim \limits_{n \to \infty} \left(\frac{n^2}{n^3} + \frac{n^2}{n^3} + \dots + \frac{n^2}{n^3}\right)$$
...but we can only let ${n \to \infty}$ and calculate the limit if all of the individual limits are of defined form (is this correct?). That would be the case here, so we have:
$= \dots \lim \limits_{n \to \infty} \left(\frac{1}{n} + \frac{1}{n} + \dots + \frac{1}{n}\right) =$[ letting ${n \to \infty}]$ $= 0 + 0 + \dots + 0 = 0$
and the results we get are not the same.
Where did I go wrong? 
 A: The problem you have described is common and there are three possible scenarios to cover:

*

*Number of terms is finite and independent of $n$: The limit of a sum is equal to sum of limits of terms provided each term has a limit.

*Number of terms is infinite: Some sort of uniform convergence of the infinite series is required and under suitable conditions the limit of a sum is equal to the sum of limits of terms.

*Number of terms is dependent on $n$: This is the case which applies to the question at hand. The problem is difficult compared to previous two cases and a partial solution is provided by Monotone Convergence Theorem. Sadly the theorem does not apply to your specific example. But it famously applies to the binomial expansion of $(1+n^{-1})^n$ and gives the result $$e=1+1+\frac{1}{2!}+\frac{1}{3!}+\dots$$
A: Because the number of terms goes up exactly as the size of each term goes down.
Specifically $$\lim \limits_{n \to \infty} \Big(\underbrace{\frac{1}{n} + \frac{1}{n} + \dots + \frac{1}{n}}_{n\text{ times}}\Big) = \lim \limits_{n \to \infty} \sum_{i=1}^n \frac 1n$$
Does that help?
