When is the limit of a sum equal to the sum of limits?
Solution 1:
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?
Solution 2:
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 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 $$e=1+1+\frac{1}{2!}+\frac{1}{3!}+\dots$$