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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?

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Say either "the sequence converges to" or "the limit is". –  lhf Jun 23 '12 at 11:52
    
@lhf Right, thanks! –  dtldarek Jun 23 '12 at 11:54
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2 Answers 2

up vote 8 down vote accepted

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 $$ \lim_{n\to \infty}g_n=2\int_0^1f(x)dx=2\int_0^{\pi/2}\frac{\cos t}{\sqrt{1-\sin^2t}}dt=2\int_0^{\pi/2}dt=\pi. $$

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Thank you, that helped me a lot! –  dtldarek Jun 23 '12 at 14:02
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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

$$ 2 \int_{0}^{1} \frac{dx}{\sqrt{2x - x^2}} = \pi.$$

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Thank you too, even if Mercy's answer is more appealing to me :-) –  dtldarek Jun 23 '12 at 14:04
    
@dtldarek, I also agree. My answer is rather a sketch. –  sos440 Jun 23 '12 at 14:15
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