Prove that every converging limit $\lim_{n \to \infty} \sum_{k=1}^{a(n)} f(k,n)$ is essentially a riemann sum. Let $a(n)$ be a strictly increasing function of $n$.
Proof that every converging limit 
$$\lim_{n \to \infty} \sum_{k=1}^{a(n)} f(k,n)$$
is essentially a Riemann sum.
 A: Providing a complete answer requires clarity on what is meant by "essentially" a Riemann sum and, most likely, narrowing the class of functions under consideration. Without making that attempt, I think examining a few cases might shed some light on whether this question can be answered in the most general form.
I am speculating you mean that computation of the limit can be reduced to the evaluation of a definite Riemann integral, either (1) directly as the limit of a Riemann sum or (2) in a more general limiting process where one step involves the limit of a Riemann sum.
As an example of category (1), consider $a(n) = n$ and $f(k,n) = k/(n^2 + k^2)$ where we have a bona fide Riemann sum,
$$\lim_{n \to \infty}\sum_{k=1}^n \frac{k}{n^2 + k^2} = \lim_{n \to \infty} \frac{1}{n}\sum_{k=1}^n \frac{k/n}{1 + (k/n)^2} = \int_0^1 \frac{x}{1 + x^2} \, dx = \frac{\log 2}{2} $$
With a slight modification, where $a(n) = n$ and $f(k,n) = k/(n^2 + k),$ we have an example from category (2), 
$$\lim_{n \to \infty}\sum_{k=1}^n \frac{k}{n^2 + k} = \lim_{n \to \infty} \frac{1}{n}\sum_{k=1}^n \frac{k/n}{1 + k/n^2} $$
In this case, we no longer have a Riemann sum.  The limit can be shown to be $1/2\,$ by various means. However, we can reintroduce Riemann sums by evaluating as an iterated  limit where the inner limit involves the sum,
$$\begin{align} \lim_{n \to \infty}\sum_{k=1}^n \frac{k}{n^2 + k} &= \lim_{m \to \infty} \lim_{n \to \infty} \frac{1}{n}\sum_{k=1}^n \frac{k/n}{1 + (k/n)(1/m)} \\ &=  \lim_{m \to \infty} \int_0^1 \frac{x}{1 + x/m} \, dx  \\ &= \int_0^1 x \, dx  \\ &= \frac{1}{2} \end{align}.$$
The steps of conversion to a double limit and passing the limit under the integral can be justified in this case using uniform and dominated convergence, respectively. 
This raises the question of what general characteristics of $f$  where $\sum f(k,n)$ converges would permit such steps.  I suspect that if $f$ is monotone, then this may work in many more cases. 
Finally, we should address the general form for the upper limit $a(n)$ which you specify as strictly increasing and, presumably, tending to $+\infty$.  Here we find problems arising in terms of uniqueness.
An example is,
$$\lim_{n \to \infty} \sum_{k=1}^{a(n)} \frac{n}{n^2 + k^2} = \lim_{n \to \infty} \frac{1}{n}\sum_{k=1}^{a(n)} \frac{1}{1 + (k/n)^2} \\ = \begin{cases} 0, \,\,\,\,\,\,\,\,\,\,\,\, a(n) = \sqrt{n} \\ \pi/4, \,\,\,\,\,a(n) = n\\ \pi/2, \,\,\,\,\, a(n) = n^2\end{cases}.$$
In the case of $a(n) = n^2,$ this becomes an improper Riemann integral.
