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I found that the series $$s(n) = \frac{2}{n} \cdot \sum_{i=1}^n \sqrt{\frac{n}{i-\frac{1}{2}}-1}$$ converges to $\pi$ as $n \to \infty$. To verify this I have computed some values:

$n$ $s(n)$
$10^1$ 2.76098
$10^3$ 3.10333
$10^5$ 3.13776
$10^6$ 3.14038
$10^7$ 3.14121

Which seems to support the claim, however, this is no proof of the convergence.

I would not know how to begin on a proof of this limit and did not find any similar formula in known approximation formulas.

Does anyone have an idea on how such a proof can be constructed?

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    $\begingroup$ Amazing discovery! I am guessing that to do this, we need to turn the sum into a product using logarithm properties. $\endgroup$ Jan 24, 2023 at 16:48
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    $\begingroup$ Maybe you can express it as a Riemann sum $\endgroup$ Jan 24, 2023 at 16:55
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    $\begingroup$ These are some fascinating results, and the proofs below are just as impressive. I come to wonder, how did you come up with this? Did you randomly just stumble upon it? Or was there some inspiration? $\endgroup$
    – bob
    Jan 26, 2023 at 22:46
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    $\begingroup$ @bob, I believe I was checking out some constants in the OEIS (no idea which one) and when I made some alterations to a series I recognized the converging number to be $\pi/2$. I thought that the series already had been found as I derived it from a couple of others and many series have been found to approximate $\pi$, so I wrote it on my to-do list (luckily) and laid it aside for about a year or two. Recently, crossing off this long-standing to-do, I decided to write this post. $\endgroup$ Jan 26, 2023 at 23:27
  • $\begingroup$ @jorisperrenet, that's very cool to hear. I also recently tried making a limit expression equal to $\pi$ here math.stackexchange.com/q/4622218/1141581, but it didn't turn out as clean as yours did. I will look into some constants in the OEIS to maybe see some nice results. Thanks for the reply. $\endgroup$
    – bob
    Jan 27, 2023 at 5:09

4 Answers 4

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$\newcommand{\d}{\,\mathrm{d}}$Note that: $$\frac{1}{n}\sum_{k=1}^n\sqrt{\frac{1}{k/n-1/2n}-1}\to\int_0^1\sqrt{x^{-1}-1}\d x=\frac{\pi}{2}$$

Substituting $x=t^{-1},t=u+1,u=w^2$ in that order equates the integral with: $$2\int_0^\infty\frac{w^2}{(w^2+1)^2}\d w$$And this is manageable.

Or, let $x=\sin^2t$. You then deal with: $$\int_0^{\pi/2}2\sin(t)\cos(t)\cot(t)\d t=2\int_0^{\pi/2}\cos^2t\d t$$Which is even easier.


To justify treating the partial sums as Riemann sums, it is sufficient to demonstrate: $$\lim_{n\to\infty}S_n=\lim_{n\to\infty}\left[\left(\frac{1}{n}\sum_{k=1}^n\sqrt{\frac{n}{k-\frac{1}{2}}-1}\right)-\left(\frac{1}{n}\sum_{k=1}^n\sqrt{\frac{n}{k}-1}\right)\right]=0$$

I present a proof of this by elementary bounds. No need for fancy convergence theorems here!

Using the difference of two squares we can bound, for every $k\ge1$ and every $n>k$: $$\begin{align}0&\le\sqrt{\frac{n}{k-\frac{1}{2}}-1}-\sqrt{\frac{n}{k}-1}\\&=\frac{\frac{\frac{1}{2n}}{\frac{k}{n}\left(\frac{k}{n}-\frac{1}{2n}\right)}}{\sqrt{\frac{n}{k-\frac{1}{2}}-1}+\sqrt{\frac{n}{k}-1}}\\&\le\frac{1}{2}\sqrt{\frac{k}{n-k}}\cdot\frac{n}{k(2k-1)}\\&=\frac{1}{2}\frac{\sqrt{n}}{(2k-1)\sqrt{k(1-k/n)}}\\&\le\frac{\sqrt{n}}{2k\sqrt{k(1-k/n)}}\end{align}$$Summing in $k$ and dividing by $n$ finds: $$0\le S_n\le\frac{1}{n\sqrt{2n-1}}+\frac{1}{2\sqrt{n}}\sum_{k=1}^{n-1}\frac{1}{k\sqrt{k(1-k/n)}}$$

Estimating the sum explicitly is a little fiddly. The map $x\mapsto\frac{1}{x\sqrt{x(1-x/n)}}$ is convex. It is initially decreasing but then increases after $x=3n/4$. If $m$ is the floor of $3n/4$ and $n$ is considered large, we can put: $$\begin{align}\sum_{k=1}^{n-1}\frac{1}{k\sqrt{k(1-k/n)}}&=\sum_{k=1}^m\frac{1}{k\sqrt{k(1-k/n)}}+\sum_{k=m+1}^{n-1}\frac{1}{k\sqrt{k(1-k/n)}}\\&\le\frac{1}{\sqrt{1-1/n}}+\int_1^m\frac{1}{x\sqrt{x(1-x/n)}}\d x\\&+\int_{m+1}^{n-1}\frac{1}{x\sqrt{x(1-x/n)}}\d x+\frac{1}{(n-1)\sqrt{1-1/n}}\\&<(1-1/n)^{-3/2}+\int_1^{n-1}\frac{1}{x\sqrt{x(1-x/n)}}\d x\\&=(1-1/n)^{-3/2}+\frac{2(n-2)}{\sqrt{n(n-1)}}\end{align}$$

Overall I get: $$0<S_n<\frac{1}{n\sqrt{2n-1}}+\frac{n}{2(n-1)\sqrt{n-1}}+\frac{n-2}{n\sqrt{n-1}}$$When $n$ is large enough that $m<n-2$, e.g. this is for sure when $n>12$.

By the squeeze theorem, it is now clear that $S_n$ vanishes - as desired.

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    $\begingroup$ I don't understand why this was downvoted, as it clearly answers the question, so I am upvoting it back. $\endgroup$ Jan 24, 2023 at 16:55
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    $\begingroup$ @KamalSaleh I suppose the initial revision was a little terse. I've fleshed out some details now : ) $\endgroup$
    – FShrike
    Jan 24, 2023 at 17:44
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    $\begingroup$ Using this same proof it seems to me that $\frac{2}{n} \cdot \sum_{i=1}^n \sqrt{\frac{n}{i+a}-1}$ converges to $\pi$ as $n \to \infty$ for all $a \in \mathbb R \setminus \mathbb N_{\geq 1}$, a great result. $\endgroup$ Jan 24, 2023 at 18:50
  • $\begingroup$ @FShrike why did you delete your question on the residues? I'm interested in were you went wrong $\endgroup$
    – Fix
    Feb 12, 2023 at 23:22
  • $\begingroup$ @Fix My expression for the Laurent coefficients was correct. But, my counterexample was wrong! So there isn’t really a question anymore $\endgroup$
    – FShrike
    Feb 13, 2023 at 9:44
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To the nice solution by @FShrike we can also try to find next asymptotic terms.

Denoting $\,\displaystyle S(n)=\frac{2}{n}\sum_{k=1}^n\sqrt{\frac{n}{k-\frac{1}{2}}-1}$ $$S(n)=\frac{2}{n}\sum_{k=1}^n\sqrt\frac{n}{k-\frac{1}{2}}+\frac{2}{n}\sum_{k=1}^n\frac{\sqrt{n-k+\frac{1}{2}}-\sqrt n}{\sqrt{k-\frac{1}{2}}}$$ $$=\frac{2}{n}\sum_{k=1}^n\sqrt\frac{n}{k-\frac{1}{2}}-\frac{2}{n}\sum_{k=1}^n\frac{\sqrt{k-\frac{1}{2}}}{\sqrt{n-k+\frac{1}{2}}+\sqrt n}=S_1+S_2\tag{1}$$ where $$S_1=\frac{2\sqrt2}{\sqrt n}\sum_{k=1}^n\Big(\frac{1}{\sqrt{2k-1}}+\frac{1}{\sqrt{2k}}-\frac{1}{\sqrt{2k}}\Big)$$ $$=\frac{2\sqrt2}{\sqrt n}\sum_{k=n+1}^{2n}\frac{1}{\sqrt k}+\frac{2\sqrt2}{\sqrt n}\Big(1-\frac{1}{\sqrt2}\Big)\sum_{k=1}^n\frac{1}{\sqrt k}\tag{2}$$ Using the Euler-MacLaurin formula, $$\frac{2\sqrt2}{\sqrt n}\sum_{k=n+1}^{2n}\frac{1}{\sqrt k}=\frac{2\sqrt2}{\sqrt n}\int_{n+1}^{2n}\frac{dk}{\sqrt k}+\frac{2\sqrt2}{\sqrt n}\frac{1}{2}\Big(\frac{1}{\sqrt k}\,\bigg|_{k=n+1}+\frac{1}{\sqrt k}\,\bigg|_{k=2n}\Big)+O(n^{-3/2})$$ $$=\frac{4\sqrt2}{\sqrt n}\Big(\sqrt{2n}-\sqrt{n+1}\Big)+\frac{\sqrt2}{\sqrt n}\Big(\frac{1}{\sqrt{2n}}+\frac{1}{\sqrt{n+1}}\Big)+O(n^{-3/2})$$ $$=4\sqrt2(\sqrt 2-1)-\frac{2\sqrt2}{n}+\frac{1+\sqrt2}{n}+O(n^{-3/2})\tag{3}$$ Using the presentation of zeta-function $$ \zeta\Big(\frac{1}{2}\Big)=\sum_{k=1}^n\frac{1}{\sqrt k}-2\sqrt n-\frac{1}{2}\frac{1}{\sqrt n}+O(n^{-3/2})\tag{4}$$ and putting (3) and (4) into (2) $$S_1=4+\frac{2(\sqrt 2-1)}{\sqrt n}\zeta\Big(\frac{1}{2}\Big)+O(n^{-3/2})\tag{5}$$ Using again the Euler-Maclaurin formula we notice that we only need to take two first terms to evaluate $S_2$ with the required accuracy.

Therefore, $$S_2=-\frac{2}{n}\sum_{k=1}^n\frac{\sqrt{k-\frac{1}{2}}}{\sqrt{n-k+\frac{1}{2}}+\sqrt n}=-\frac{2}{n}\int_1^n\frac{\sqrt{k-\frac{1}{2}}}{\sqrt{n-k+\frac{1}{2}}+\sqrt n}\,dk$$ $$-\,\frac{2}{n}\frac{1}{2}\bigg(\frac{\sqrt{k-\frac{1}{2}}}{\sqrt{n-k+\frac{1}{2}}+\sqrt n}\,\bigg|_{k=1}+\frac{\sqrt{k-\frac{1}{2}}}{\sqrt{n-k+\frac{1}{2}}+\sqrt n}\,\bigg|_{k=n}\bigg)+O(n^{-3/2})$$ $$=-2\int_\frac{1}{2n}^{1-\frac{1}{2n}}\frac{\sqrt t}{1+\sqrt{1-t}}dt-\frac{1}{n}+O(n^{-3/2})$$ $$=\pi-4+\frac{1}{n}-\frac{1}{n}+O(n^{-3/2})=\pi-4+O(n^{-3/2})\tag{6}$$ Putting in turn (5) and (6) into (1) $$\boxed{\,\,S(n)=\pi+\frac{2(\sqrt2-1)}{\sqrt n}\zeta\Big(\frac{1}{2}\Big)+O(n^{-3/2})\,\,}$$ The numeric check with WolframAlpha confirms the answer: $$n=100\,\quad S(n)=3.02067..\quad\text{approximation}=3.020612...$$ $$n=1000\quad S(n)=3.10334..\quad\text{approximation}=3.103355...$$

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    $\begingroup$ This is pure beauty ! $\endgroup$ Jan 25, 2023 at 3:25
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    $\begingroup$ @Svyatoslav I derived the higher terms in your asymptotics. See my answer. $\endgroup$
    – Gary
    Jan 26, 2023 at 11:12
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I shall derive a complete asymptotic expansion for $s(n)$. The argument is analogous to the one given here. First note that \begin{align*} \frac{n}{2}s(n) & = \sum\limits_{i = 1}^n {\sqrt {\frac{n}{{i - 1/2}} - 1} } = \sum\limits_{i = 1}^n {\sqrt {\frac{{2n - 2i + 1}}{{2i - 1}}} } \\ & = [x^{2n} ]\left( {\sum\limits_{k = 1}^\infty {\frac{{x^{2k} }}{{\sqrt {2k - 1} }}} } \right)\left( {\sum\limits_{k = 0}^\infty {\sqrt {2k + 1} x^{2k} } } \right) \\ & = [x^{2n - 1} ]\left( {\chi _{1/2} (x)\frac{{{\rm d}\chi _{1/2} (x)}}{{{\rm d}x}}} \right) = \frac{1}{2}[x^{2n - 1} ]\frac{{{\rm d}\chi _{1/2}^2 (x)}}{{{\rm d}x}} \\ & = \frac{1}{2}\frac{1}{{(2n - 1)!}}\left[ {\frac{{{\rm d}^{2n} \chi _{1/2}^2 (x)}}{{{\rm d}x^{2n} }}} \right]_{x = 0} = \frac{n}{{2\pi {\rm i}}}\oint_{(0 + )} {\frac{{\chi _{1/2}^2 (z)}}{{z^{2n + 1} }}{\rm d}z} , \end{align*} where $\chi$ is the Legendre $\chi$-function and $[x^n]$ is the $n$th coefficient extraction operator. Therefore, $$ s(n) = \frac{1}{{\pi {\rm i}}}\oint_{(0 + )} {\frac{{\chi _{1/2}^2 (z)}}{{z^{2n + 1} }}{\rm d}z} $$ for any $n\ge 1$. The function $\chi _{1/2}^2(z)$ is analytic on $\mathbb{C}\setminus ((-\infty,-1] \cup [1,+\infty))$ and is $\mathcal{O}(\log z)$ for large $|z|$. Therefore, we can blow up the contour of integration to arrive at $$ s(n) = \frac{1}{{\pi {\rm i}}}\int_{\mathscr{H}^-} {\frac{{\chi _{1/2}^2 (z)}}{{z^{2n + 1} }}{\rm d}z} + \frac{1}{{\pi {\rm i}}}\int_{\mathscr{H}^+} {\frac{{\chi _{1/2}^2 (z)}}{{z^{2n + 1} }}{\rm d}z} , $$ where the contours $\mathscr{H}^-$, $\mathscr{H}^+$ are Hankel contours surrounding $(-\infty,-1]$ and $[1,+\infty)$ in the clockwise sense. Since $\chi _{1/2}^2(z)$ is even, we can simplify the above to $$ s(n) = \frac{2}{{\pi {\rm i}}}\int_{\mathscr{H}^+} {\frac{{\chi _{1/2}^2 (z)}}{{z^{2n + 1} }}{\rm d}z}. $$ Substituting $z=\mathrm{e}^w$, $$ s(n) = \frac{2}{{\pi {\rm i}}}\int_\mathscr{H} {{\rm e}^{ - 2nw} \chi _{1/2}^2 ({\rm e}^{w} ){\rm d}w} $$ where $\mathscr{H}$ is a Hankel contour surrounding the positive real line in the clockwise sense. Collapsing the contour onto the two sides of the positive real axis and extracting the contribution coming from the origin, we find $$ s(n) = \pi + \int_0^{ + \infty } {{\rm e}^{ - 2nt} f(t){\rm d}t} $$ with $$ f(t): = \mathop {\lim }\limits_{\varepsilon \to 0^ + } \frac{2}{{\pi {\rm i}}}\left[ {\chi _{1/2}^2 ({\rm e}^{t + {\rm i}\varepsilon } ) - \chi _{1/2}^2 ({\rm e}^{t - {\rm i}\varepsilon } )} \right]. $$ The contribution from the origin follows from the series expansion $$ \chi _{1/2} ({\rm e}^w ) = \frac{1}{2}\sqrt {\frac{\pi }{{ - w}}} + \sum\limits_{k = 0}^\infty {(1 - 2^{k - 1/2} )\zeta (1/2 - k)\frac{{w^k }}{{k!}}} $$ valid for $0<|w|<\pi$. This may be deduced from the relation $$ \chi _{1/2} ({\rm e}^w ) = \operatorname{Li}_{1/2} ({\rm e}^w ) - \frac{1}{{\sqrt 2 }}\operatorname{Li}_{1/2} ({\rm e}^{2w} ) $$ with the polylogarithm and the analogous series expansion of the latter. From this series it is readily found that $$ f(t) = \frac{4}{{\sqrt \pi }}\sum\limits_{k = 0}^\infty {(1 - 2^{k - 1/2} )\zeta (1/2 - k)\frac{{t^{k - 1/2} }}{{k!}}} $$ for $0<|t|<\pi$. Hence, by Watson's lemma, $$\boxed{ s(n) \sim \pi + 2\sum\limits_{k = 0}^\infty {\binom{k-1/2}{k}(2^{1/2 - k} - 1)\zeta (1/2 - k)\frac{1}{{n^{k + 1/2} }}} } $$ as $n\to +\infty$.

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    $\begingroup$ I had a strong suspicion that the asymptotics would be similar to the one of the previous problem - the series of zeta functions, and that there would be no integer powers of 1/n (the first power even fell out of my solution). But my level was not even close to allowing me to develop this topic; I can only watch in fascination at what depths can be reached by a professional who knows how to use the most powerful mathematical tools. It all looks a lot like magic :) $\endgroup$
    – Svyatoslav
    Jan 26, 2023 at 11:39
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    $\begingroup$ @Svyatoslav I am always admiring yours and Gary's answers. Of the three analyses of this series, mine was by far the most trivial ;) $\endgroup$
    – FShrike
    Jan 30, 2023 at 18:39
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Consider the sum $$\frac{1}{n}\sum_{k=1}^{n}f\left(\frac{2k-1}{2n}\right),$$ where $f$ is decreasing on $(0,1]$. If $\int_{0}^{1}f(x)\,{\rm d}x$ is convergent or is Riemann integral, then $$\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^{n}f\left(\frac{2k-1}{2n}\right) =\int_{0}^{1}f(x)\,{\rm d}x.$$

Proof: For $1\leq k\leq n-1$, we have $$\frac{1}{n}f\left(\frac{2k+1}{2n}\right) \leq\int_{\frac{2k-1}{2n}}^{\frac{2k+1}{2n}}f(x)\,{\rm d}x \leq f\left(\frac{2k-1}{2n}\right).$$ $$\frac{1}{n}f\left(\frac{1}{2n}\right) \leq\int_{0}^{\frac{1}{2n}}f(x)\,{\rm d}x +\frac{1}{2n}f\left(\frac{1}{2n}\right),$$ $$\frac{1}{2n}f\left(\frac{2n-1}{2n}\right) +\int_{\frac{2n-1}{2n}}^{1}f(x)\,{\rm d}x \leq\frac{1}{n}f\left(\frac{2n-1}{2n}\right).$$ By the above inequalities, we get $$\frac{f\left(\frac{2n-1}{2n}\right)}{2n}+\int_{\frac{1}{2n}}^1f(x)\,{\rm d}x \leq\frac{1}{n}\sum_{k=1}^{n}f\left(\frac{2k-1}{2n}\right) \leq\int_{0}^{\frac{2n-1}{2n}}f(x)\,{\rm d}x+\frac{f\left(\frac{1}{2n}\right)}{2n}.\tag{1}$$ It is easy to see that $$\lim_{n\to\infty}\left[\frac{1}{2n}f\left(\frac{2n-1}{2n}\right)+\int_{\frac{1}{2n}}^1f(x)\,{\rm d}x\right]=\int_{0}^{1}f(x)\,{\rm d}x;\tag{2}$$ $$\lim_{n\to\infty}\left[\int_{0}^{\frac{2n-1}{2n}}f(x)\,{\rm d}x+\frac{1}{2n}f\left(\frac{1}{2n}\right)\right]=\int_{0}^{1}f(x)\,{\rm d}x.\tag{3}$$ By $(1),(2),(3)$, we have $$\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^{n}f\left(\frac{2k-1}{2n}\right) =\int_{0}^{1}f(x)\,{\rm d}x.$$

Choose $$f(x)=2\sqrt{\frac{1}{x}-1},\quad x\in(0,1],$$ by the above result, we get $$\lim_{n\to\infty}\frac{2}{n}\sum_{i=1}^n \sqrt{\frac{n}{i-\frac{1}{2}}-1} =\lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^n2\sqrt{\frac{1}{\frac{2i-1}{2n}}-1} =2\int_{0}^{1}\sqrt{\frac{1}{x}-1}\,{\rm d}x=\pi.$$

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