Limit in Infinite Summation Over Function in Two Variables When can the following statement be made?
$$\lim_{N\to \infty} \frac{1}{N}\sum_{n\leq N}\lim_{k\to \infty} f(n,k)=\lim_{N\to \infty} \lim_{k\to \infty} \frac{1}{N}\sum_{n\leq N} f(n,k)= \lim_{k\to \infty}\lim_{N\to \infty} \frac{1}{N}\sum_{n\leq N} f(n,k)$$
Additionally, under what conditions can I let $N$ and $k$ tend to infinity at the same time at obtain
$$\lim_{N\to \infty} \frac{1}{N}\sum_{n\leq N} f(n,N)$$
assuming of course that the limit exists?
 A: In general for double sequences, a sufficient condition for 
$$\lim_{N \to \infty} (\lim_{k \to \infty} S_{Nk}) = \lim_{k \to \infty} (\lim_{N \to \infty} S_{Nk}) = \lim_{N \to \infty} S_{NN},$$
is the existence of the limit $\lim_{N \to \infty} S_{Nk} $ for all k and the uniform convergence of $S_{Nk} \to L_N$ as $k \to \infty$ for all $N$ or vice versa.
Sketch of proof:
(1) Using the uniform convergence, it can be shown that $L_N$ is a Cauchy sequence and, hence, converges to some limit $L$.
(2) It can then be shown that the double sequence converges in the sense that given any $\epsilon > 0$ there exists $M$ such that if $N,k > M$ we have $|S_{Nk} - L| < \epsilon.$
(3) From (2) it follows that both iterated limits are equal to $L$ and also equal to the limit of $S_{NN}$.
That $\lim_{N \to \infty} S_{NN} = L$ follows directly from (2) with $k = N$.
Again from (2), if $|S_{Nk} - L| < \epsilon$ for sufficiently large $N$ and $k$ , then  we have $\lim_{k \to \infty}|S_{Nk} - L| = | \lim_{k \to \infty}S_{N,k} - L| \leqslant \epsilon$, which implies $\lim_{N \to \infty} \lim_{k \to \infty} S_{Nk} = L$.
And so forth ...
