Limit of a sum of two variables Recently at my calculus course we are doing derivatives and integrals. 

I've stumbled upon a sum that seems to have nothing in common with our current objectives, though I'm sure it does have, but still I have no idea how to solve it. 

$$ \lim_{n\to\infty} \sum_{k=1}^{n} \frac{1}{4n - \frac{k^2}{n}} $$

I tried to represent it as a $$ f_{n}(x) = \sum_{k=1}^{n} \frac{x}{4n - \frac{k^2}{n}}$$ and to resolve it at a point $x = 1$, but I failed because of the two variables instead of one.

EDIT:

Of course, I would like to get some CLUES how to solve it, not the whole solution.
Observe that we have
$$
\int_0^1\frac1{4-x^2}dx=\lim_{n\to\infty} \frac1n\sum_{k=1}^n\frac1{4-\frac{k^2}{n^2}}
$$
by definition of the integral.
Let $f(t)=\frac{1}{4-t^2} $ then $$\frac{1}{4n -\frac{k^2}{n}} =\frac{1}{4-\left(\frac{k}{n}\right)^2} \cdot \left(\frac{k}{n} -\frac{k-1}{n}\right) =f\left(\frac{k}{n} \right)\cdot \left(\frac{k}{n} -\frac{k-1}{n}\right)$$
Hence 
$$\lim_{n\to\infty} \sum_{k=1}^n \frac{1}{4n -\frac{k^2}{n}} =\lim_{n\to\infty} \sum_{k=1}^n f\left(\frac{k}{n} \right)\cdot \left(\frac{k}{n} -\frac{k-1}{n}\right)= \int_0^1 f(v)dv =\ln \sqrt[8]{\frac{5}{3}}$$