<s>[INST] $ \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???
 [/INST] Computation error:
\begin{align}
\int_0^1 \frac{dx}{4-x^2} &= \frac{1}{4}\int_0^1 \frac{1}{2+x} + \frac{1}{2-x}\,dx\\
&= \frac{1}{4} \bigl[\log (2+x) - \log (2-x)\bigr]_0^1\\
&= \frac{1}{4}\bigl[\log 3 - \log 1 - \log 2 + \log 2\bigr]\\
&= \frac{1}{4}\log 3\\
&> \frac{1}{4}.
\end{align}
</s>