Evaluate $\lim \limits_{n \to \infty} \sum_{k=1}^{\lfloor n/2 \rfloor} \frac {n^2}{n^2-k^2}$ To evaluate the limit $\lim \limits_{n \to \infty} \sum_{k=1}^{\lfloor n/2 \rfloor} \frac {n^2}{n^2-k^2}$  I want to use Riemann sums and integrate but I'm having a bit of a hard time as I'm not sure what partition to use. I figured that the relevant function would be $f(x) = \frac{1}{1-x^2}$ but I can't see how to get the partition factor in the mix.
 A: Actually any term of the sum is $> 1$ and the sequence goes to infinity which is a bit weird here if you are thinking to a Riemann sum. Are you sure that the numerator is $n^2$? Maybe it is $n$.
Note that for $k=1,\dots, \lfloor n/2 \rfloor$,
we have that 
$$0<\frac{1}{n}<\frac{2}{n}<\dots <\frac{k}{n} <\dots<\frac{\lfloor n/2 \rfloor}{n}\leq \frac{1}{2}.$$
Moreover, if the numerator is $n$ then, as $n$ goes to infinity,
$$ \sum_{k=1}^{\lfloor n/2 \rfloor} \frac {n}{n^2-k^2}=\left(\frac{1}{n}\sum_{k=1}^{\lfloor n/2 \rfloor} f(k/n)\right)\to \int_0^{1/2}f(x)dx$$
with $f(x) = \frac{1}{1-x^2}$. Can you take it from here?
A: Since
$$
\begin{align}
\sum_{k=1}^{\lfloor n/2\rfloor}\frac{n^2}{n^2-k^2}
&=\sum_{k=1}^{\lfloor n/2\rfloor}\frac1{1-\frac{k^2}{n^2}}\\
&\ge\lfloor n/2\rfloor
\end{align}
$$
Thus, the limit is infinity.

However, if the question had been
$$
\begin{align}
\lim_{n\to\infty}\sum_{k=1}^{\lfloor n/2\rfloor}\frac{n}{n^2-k^2}
&=\lim_{n\to\infty}\sum_{k=1}^{\lfloor n/2\rfloor}\frac1{1-\frac{k^2}{n^2}}\frac1n\\
&=\int_0^{1/2}\frac{\mathrm{d}x}{1-x^2}\\
&=\frac12\int_0^{1/2}\left(\frac1{1-x}+\frac1{1+x}\right)\mathrm{d}x\\
&=\frac12\left[\log\left(\frac{1+x}{1-x}\right)\right]_0^{1/2}\\[6pt]
&=\frac12\log(3)
\end{align}
$$
