Is there a (basic) way to understand why a limit cannot be brought inside of a sum, in certain situations? I have seen that the method of "dominated convergence" implies that sometimes a limit cannot be brought inside of a sum, but is there a more basic way to understand this, preferably without using integrals? 
A conceptual or simple answer would be helpful. Thank you!
 A: Using Iverson brackets, we can write the indicator function $1_{k=n}=[k=n]$.
$$
1=\lim_{n\to\infty}\overbrace{\sum_{k=1}^\infty[k=n]}^1\ne\sum_{k=1}^\infty\overbrace{\vphantom{\sum_{k=1}^\infty}\lim_{n\to\infty}[k=n]}^0=0
$$
A: Basic example helps:
$$\frac12=\lim_{x\to-1^+}\frac1{1-x}=\lim_{x\to-1^+}\sum_{n=0}^\infty x^n\ne\sum_{n=0}^\infty\lim_{x\to-1^+}x^n=\sum_{n=0}^\infty(-1)^n$$
A: A sum like $$f(t) = \sum_{i = 0}^\infty t^i$$ is not actually a sum, but is shorthand for $$f(t) = \lim_{j \to \infty} \sum_{i = 0}^{j} t^i$$ so it's actually a limit of partial sums. If I now want to take some limit with respect to $t$, I'll end up with one limit inside another, and limits in general do not commute, as one can observe by considering $e^{x - y}$ as $x, y$ go to infinity: depending on the order you do these you can get 0 or $\infty$. 
A: I think a counterexample like this one will probably best illustrate why this can happen.
Let $f_n$ be functions defined on $[0,1]$ such that each $f_n$ is zero on $[1/n,1]$, but nonetheless has integral $1$. For example, the graph of $f_n$ could be a very tall but narrow triangle with base $[0,1/n]$. Then the pointwise limit of $f_n(x)$ is the zero function, but its integral is $0$, not $1$.
