Does this sequence converge to $\pi$? I have a problem with the following sequence
$$ \lim_{n \to \infty} g_n \stackrel{?}{=} \pi $$
where
$$g_n = \sum_{k=1}^{n-1} \frac{\sqrt{\frac{2n}{k}-1}}{n-k}
 + \sum_{k=n+1}^{2n-1}\frac{\sqrt{\frac{2n}{k}-1}}{n-k}.$$
Does it converge to $\pi$? I tested experimentally that it
does, but I was unable to prove it by hand. Could anybody help,
or offer some methods of approach?
 A: Setting $i=n-k$ and $j=k-n$ we have
\begin{eqnarray}
g_n&=&
\sum_{i=1}^{n-1}\frac{1}{i}\sqrt{\frac{2n-n+i}{n-i}}-\sum_{j=1}^{n-1}\frac{1}{j}\sqrt{\frac{2n-n-j}{n+j}}\cr
&=&\sum_{i=1}^{n-1}\frac{1}{i}\sqrt{\frac{n+i}{n-i}}-\sum_{j=1}^{n-1}\frac{1}{j}\sqrt{\frac{n-j}{n+j}}\cr
&=&\sum_{k=1}^{n-1}\frac{1}{k}\left(\sqrt{\frac{n+k}{n-k}}-\sqrt{\frac{n-k}{n+k}}\right)\cr
&=&\sum_{k=1}^{n-1}\frac{1}{k}\frac{(n+k)-(n-k)}{\sqrt{n^2-k^2}}\cr
&=&2\sum_{k=1}^{n-1}\frac{1}{\sqrt{n^2-k^2}}=\frac{2}{n}\sum_{k=1}^{n-1}\frac{1}{\sqrt{1-(k/n)^2}}=-\frac{2}{n}+\frac{2}{n}\sum_{k=0}^{n-1}f(k/n),
\end{eqnarray}
with $f(x)=1/\sqrt{1-x^2}$.
Therefore
$$
\lim_{n\to \infty}g_n=2\int_0^1f(x)dx=2\int_0^{\pi/2}\frac{\cos t}{\sqrt{1-\sin^2t}}dt=2\int_0^{\pi/2}dt=\pi.
$$
A: Let $l = 2n - k$. Then by simple algebraic manipulation, we have
$$ \sum_{k=n+1}^{2n} \frac{\sqrt{\frac{2n}{k} - 1}}{n-k} = - \sum_{l=1}^{n-1} \frac{1}{(n-l)\sqrt{\frac{2n}{l} - 1}}. $$
Thus replacing the summation index by $k$ and letting $x_k = x_{k}^{(n)} = \frac{k}{n}$,
$$ g_n = \frac{1}{n} \sum_{k=1}^{n-1} \frac{1}{1 - x_k} \left( \sqrt{\frac{2}{x_k} - 1} - \frac{1}{\sqrt{\frac{2}{x_k} - 1}} \right) = \frac{1}{n} \sum_{k=1}^{n-1} \frac{2}{\sqrt{2x_k - x_k^2}}, $$
which converges to
$$ 2 \int_{0}^{1} \frac{dx}{\sqrt{2x - x^2}} = \pi.$$
