$ \lim_{n\to\infty} \sum_{k=1}^{n} \frac{1}{4n - \frac{k^2}{n}} $ appears to disagree with $\int_0^1 \frac{dx}{4-x^2}$ In question 1280454, t was asked how to find
$$ \lim_{n\to\infty} \sum_{k=1}^{n} \frac{1}{4n - \frac{k^2}{n}} $$
and of course, you can write this as 
$$ \lim_{n\to\infty} \frac{1}{n} \sum_{k=1}^{n} \frac{1}{4 - \frac{k^2}{n^2}}
= \int_0^1 \frac{dx}{4-x^2}
$$ by using the variable $x=\frac{k}{n}$ in a Reimann integral.
However, 
$$\int_0^1 \frac{dx}{4-x^2} = \frac18\left(\ln(5)-\ln(3) \right) \approx 0.06385$$
while each of the $n$ terms in 
$$
\sum_{k=1}^{n} \frac{1}{4n - \frac{k^2}{n}}$$ exceeds $\frac{1}{4n}$ so the sum must be greater than $\frac{1}{4}$ in all cases.  (Experimentatlly, the limit is about $0.275$)
What gives???

$$\dfrac1{4n-\dfrac{k^2}n}=\dfrac n{4n^2-k^2}=\dfrac14\dfrac{2n+k+(2n-k)}{(2n+k)(2n-k)}=\dfrac14\left[\dfrac1{2n+k}+\dfrac1{2n-k}\right]$$
Now follow Find the limit $\lim_{n\rightarrow\infty}\sum_{k=1}^{n}\frac{1}{k+n}$
