# How to find $\lim_{n\to\infty}\Big(\sum_{k=1}^n\sqrt{\frac{1}{k}-\frac{1}{n}}-\frac{\pi}{2}\sqrt n\Big)$?

I'm looking for the asymptotic of the series $$\displaystyle S(n)=\frac{1}{2}\sum_{k=1}^n\int_k^n\frac{dx}{x\sqrt{x(n-x)}}\,$$ at $$\,n\to\infty$$

The first term can be found, for example, via switching to Riemann sums, changing the order of integration, and is equal to $$\frac{\pi}{2}$$. To find the next term I performed integration and got $$S(n)=\frac{1}{n}\sum_{k=1}^{n-1}\sqrt{\frac{n}{k}-1}=\frac{1}{\sqrt n}\sum_{k=1}^{n-1}\sqrt{\frac{1}{k}-\frac{1}{n}}=\frac{1}{\sqrt n}\sum_{k=1}^{n-1}f(k)\tag{1}$$ To get the flavour of the next asymptotic term I used the Euler-Maclaurin formula $$\sum_{k=1}^{n-1}f(k)\sim\int_1^{n-1}f(k)dk+\frac{1}{2}\big(f(1)+f(n-1)\big)+\frac{1}{12}\big(f'(n-1)-f'(1)\big)+ ...\tag{2}$$ At $$n\to\infty$$ the first term (the integral) gives $$\frac{\pi}{2}\sqrt n-2$$; other terms in (2) give non-zero values at $$k=1$$. Evaluating a couple of such terms, I got $$\sum_{k=1}^{n-1}f(k)\sim\frac{\pi}{2}\sqrt n-2+\frac{1}{2}+\frac{1}{24}=\frac{\pi}{2}\sqrt n-1.4583$$ The numeric evaluation at WolframAlpha for $$n=1000$$ gives $$\sum_{k=1}^{n-1}f(k)-\frac{\pi}{2}\sqrt n\,\bigg|_{n=1000}=-1.46046$$ All this strongly resembles $$\displaystyle \zeta\Big(\frac{1}{2}\Big)=-1.46035...\,\,$$; at $$\,n=10000\,$$ we get even better agreement.

Questions:

1. How can we prove that $$\,\,\displaystyle \lim_{n\to\infty}\Big(\sum_{k=1}^n\sqrt{\frac{1}{k}-\frac{1}{n}}-\frac{\pi}{2}\sqrt n\Big)=\zeta\Big(\frac{1}{2}\Big)\,\,$$?
2. Can we get next asymptotic terms (at least, several of them) analytically ?
• $\zeta(1/2)$ is undefined, isn't it? Dec 13, 2022 at 19:05
• @KamalSaleh $\zeta(s)$ is defined everywhere except for $s=1$ (as a complex function) Dec 13, 2022 at 19:11
• I belive the asymptotic expansion of $S(n)$ is $$S(n) \sim \frac{\pi }{2} + \frac{1}{{\sqrt n }}\sum\limits_{m = 0}^\infty {\binom{m - 1/2}{m}\frac{{\zeta (1/2 - m)}}{{n^m }}} .$$
– Gary
Dec 14, 2022 at 9:05
• @Gary. Could we know how you did arrive to this beautiful result ? Thanks & cheers :-) Dec 14, 2022 at 9:15
• It is based on heuristics and numerics. Looking at Sangchul Lee's answer if we expand $1/\sqrt{1-s/n}$ then $$S(n) \sim \frac{\pi }{2} + \frac{1}{{\sqrt n }}\sum\limits_{m = 0}^\infty {\binom{m - 1/2}{m}\frac{1}{{n^m }}{\left( { - \frac{1}{2}\int_0^n {\frac{{s - \left\lfloor s \right\rfloor }}{{s^{1 + 1/2 - m} }}{\rm d}s} } \right)}} .$$ Thus we need that, in some sense, $$- \frac{1}{2}\int_0^n {\frac{{s - \left\lfloor s \right\rfloor}}{{s^{1 + 1/2 - m} }}{\rm d}s} \sim \zeta \left( {\tfrac{1}{2} - m} \right).$$
– Gary
Dec 14, 2022 at 9:46

Starting from the equality $$\frac{\pi}{2}\sqrt{n} = \int_{0}^{n} \sqrt{\frac{1}{x} - \frac{1}{n}} \, \mathrm{d}x$$, we get

\begin{align*} \sqrt{n}\left( S(n) - \frac{\pi}{2} \right) &= \sum_{k=1}^{n} \sqrt{\frac{1}{k} - \frac{1}{n}} - \frac{\pi}{2}\sqrt{n} \\ &= \sum_{k=1}^{n} \sqrt{\frac{1}{k} - \frac{1}{n}} - \int_{0}^{n} \sqrt{\frac{1}{x} - \frac{1}{n}} \, \mathrm{d}x \\ &= \sum_{k=1}^{n} \int_{k-1}^{k} \biggl( \int_{x}^{k} \frac{\partial}{\partial s} \sqrt{\frac{1}{s} - \frac{1}{n}} \, \mathrm{d}s \biggr) \, \mathrm{d}x \\ &= \sum_{k=1}^{n} \int_{k-1}^{k} \biggl( \int_{k-1}^{s} \frac{\partial}{\partial s} \sqrt{\frac{1}{s} - \frac{1}{n}} \, \mathrm{d}x \biggr) \, \mathrm{d}s \\ &= -\frac{1}{2} \int_{0}^{n} \frac{s - \lfloor s \rfloor}{s^{3/2}\sqrt{1 - s/n}} \, \mathrm{d}s \\ &= -\frac{1}{2} \int_{0}^{0.2022 n} \frac{s - \lfloor s \rfloor}{s^{3/2}\sqrt{1 - s/n}} \, \mathrm{d}s + \mathcal{O}(n^{-1/2}). \end{align*}

Then by the dominated convergence theorem, this converges to

$$-\frac{1}{2} \int_{0}^{\infty} \frac{s - \lfloor s \rfloor}{s^{3/2}} \, \mathrm{d}s = \zeta\left(\frac{1}{2}\right)$$

as $$n \to \infty$$. (Note that $$\zeta(s) = -s \int_{0}^{\infty} \frac{x-\lfloor x \rfloor}{x^{1+s}} \, \mathrm{d}x$$ for $$0 < \operatorname{Re}(s) < 1$$, see the entry 25.2.8 of DLMF.)

• Thank you for your nice solution! Dec 14, 2022 at 9:43
• In line three, the inner integral in $s$ is with the range $k-1\le x\le s\le k$. In line four, you integrate in $x$ with the range $k-1\le s\le x\le k$ which seems wrong to me. It should be, I think: $$\int_{k-1}^s\frac{\partial}{\partial s}\sqrt{\frac{1}{s}-\frac{1}{n}}\,\mathrm{d}x$$On the inside Dec 14, 2022 at 11:39
• @FShrike, You are absolutely correct. Indeed we need the integral $$\int_{k-1}^{s}\mathrm{d}x=s-(k-1)=s-\lfloor s\rfloor$$ for each $k-1<s<k$. Let me fix the bounds accordingly. Dec 14, 2022 at 11:47
• I may suppose 0.2022 factor is related to the current, 2022 year AD :) Dec 14, 2022 at 12:20
• @FShrike, Indeed I simply picked a number in $(0,1)$ that looks fun and reasonably arbitrary to signal that any fixed number between 0 and 1 can be used. :) Dec 14, 2022 at 12:23

Observe that $$\frac{{{\mathop{\rm Li}\nolimits} _{1/2}^2 (z)}}{2} = \sum\limits_{n = 2}^\infty {S(n)z^n } , \quad |z|<1,$$ where $$\operatorname{Li}_s(z)$$ is the polylogarithm. (This follows by integrating $$\frac{1}{z}{\mathop{\rm Li}\nolimits} _{ - 1/2} (z){\mathop{\rm Li}\nolimits} _{1/2} (z) = {\mathop{\rm Li}\nolimits}'_{1/2} (z){\mathop{\rm Li}\nolimits} _{1/2} (z) = \frac{1}{2}({\mathop{\rm Li}\nolimits} _{1/2}^2 (z))'$$.) The asymptotics of $$S(n)$$ for large $$n$$ is controlled by the singularity of the left-hand side at $$z=1$$. The polylogarithm may be expanded into a series $${\mathop{\rm Li}\nolimits} _{1/2} (z) = \sqrt \pi ( - \log z)^{ - 1/2} + \sum\limits_{k = 0}^\infty {\frac{{\zeta (1/2 - k)}}{{k!}}\log ^k z}$$ provided $$|\log z|<2\pi$$. This gives \begin{align*} \frac{{{\mathop{\rm Li}\nolimits} _{1/2}^2 (z)}}{2} & = - \frac{\pi }{{2\log z}} + \sum\limits_{k = 0}^\infty {( - 1)^k \sqrt \pi \frac{{\zeta (1/2 - k)}}{{k!}}( - \log z)^{k - 1/2} } + F(z) \\ & = \frac{\pi }{{2(1 - z)}} + \sum\limits_{k = 0}^\infty {( - 1)^k \sqrt \pi \frac{{\zeta (1/2 - k)}}{{k!}}( - \log z)^{k - 1/2} } + G(z) \end{align*} as $$z\to 1$$, where $$F(z)$$ and $$G(z)$$ are holomorphic at $$z=1$$. Accordingly, $$\sum\limits_{n = 2}^\infty {\left( {S(n) - \frac{\pi }{2}} \right)z^n } = \sum\limits_{k = 0}^\infty {( - 1)^k \sqrt \pi \frac{{\zeta (1/2 - k)}}{{k!}}( - \log z)^{k - 1/2} } + G(z)$$ as $$z\to 1$$. The behaviour of the right-hand side near $$z=1$$ is $$\sqrt \pi \zeta (1/2)(1 - z)^{ - 1/2} - \sqrt \pi \left( {\zeta ( - 1/2) + \frac{{\zeta (1/2)}}{4}} \right)(1 - z)^{1/2} + \mathcal{O}((1 - z)^{3/2} ) + \mathcal{O}(1).$$ Singularity analysis shows that the $$n$$th Maclaurin coefficient of this function is $$\frac{{\zeta (1/2)}}{{n^{1/2} }} + \frac{{\zeta ( - 1/2)}}{{2n^{3/2} }} + \mathcal{O}\!\left( {\frac{1}{{n^{5/2} }}} \right)$$ as $$n\to +\infty$$. Consequently, $$S(n) = \frac{\pi }{2}+\frac{{\zeta (1/2)}}{{n^{1/2} }} + \frac{{\zeta ( - 1/2)}}{{2n^{3/2} }} + \mathcal{O}\!\left( {\frac{1}{{n^{5/2} }}} \right)$$ as $$n\to +\infty$$. This is in agreement with the conjectured asymptotic expansion $$S(n) \sim \frac{\pi }{2} + \frac{1}{{\sqrt n }}\sum\limits_{m = 0}^\infty {\binom{m - 1/2}{m}\frac{{\zeta (1/2 - m)}}{{n^m }}} , \quad n\to +\infty.$$ It seems that when doing the singularity analysis, many terms cancel each-other leaving us with a clean result. The proof of the general asymptotic expansion remains open for now.

• Thank you for your solution and general approach to the problem! It is fascinating for me to see how professional methods lead to the answer - some of them a beyond my abilities ) Dec 15, 2022 at 8:08
• @metamorphy Thank you. That is exactly what I tried but somehow could not get the expansion for g at the origin. I looked at $f(\mathrm{e}^t)-f(\mathrm{e}^{t+2\pi \mathrm{i}})$, which is wrong.
– Gary
Dec 15, 2022 at 22:10

We may use the integral version of Euler-Maclaurin formula to obtain an asymptotic formula with explicit big O error term. For any integer $$a and continuously differentiable $$f(x)$$ on $$[a,b]$$ there is

$$\sum_{k=a}^bf(k)=\int_a^bf(x)\mathrm dx+\frac12f(a)+\frac12f(b)+\int_a^b\overline B_1(x)f'(x)\mathrm dx,\tag1$$

where $$\overline B_n(x)$$ denotes $$n$$'th periodic Bernoulli polynomial. Therefore, we see that when $$a=1$$, $$b=n$$, and $$f_n(x)=\sqrt{x^{-1}-n^{-1}}$$ there is

$$\sum_{k=1}^nf_n(k)=\underbrace{\int_1^nf_n(x)\mathrm dx}_{S_1}+\frac12+\underbrace{\int_1^n\overline B_1(x){-x^{-2}\over2\sqrt{x^{-1}-n^{-1}}}\mathrm dx}_{S_2}+O\left(\frac1n\right).\tag2$$

For $$S_1$$, substitution $$x=nt$$ gives

\begin{aligned} n^{-1/2}S_1 &=\int_{1/n}^1\sqrt{t^{-1}-1}\mathrm dt \\ &=\int_0^1t^{1/2-1}(1-t)^{3/2-1}\mathrm dt-\int_0^{1/n}t^{-1/2}\left\{1+O(t)\right\}\mathrm dt \\ &=\frac\pi2-\int_0^{1/n}t^{-1/2}\mathrm dt+O(n^{-3/2})=\frac\pi2-2n^{-1/2}+O(n^{-3/2}), \end{aligned}

so we have

$$S_1=\frac\pi2\sqrt n-2+O\left(\frac1n\right)\tag3$$

For $$S_2$$, note that the integral of $$\overline B_1(x)$$ is bounded, so integration by parts gives

\begin{aligned} S_2 &=-\frac12\int_1^{n/2}\overline B_1(x)x^{-3/2}\left(1-\frac xn\right)^{-1/2}\mathrm dx+O(n^{-1/2}) \\ &=-\frac12\int_1^{n/2}\overline B_1(x)x^{-3/2}\left\{1+{x\over2n}+O\left(x^2\over n^2\right)\right\}\mathrm dx+O(n^{-1/2}) \\ &=-\frac12\int_1^{n/2}\overline B_1(x)x^{-3/2}\mathrm dx+O(n^{-1/2}) \\ &=-\frac12\int_1^\infty\overline B_1(x)x^{-\frac12-1}\mathrm dx+O(n^{-1/2}). \end{aligned}

If we plug $$a=1$$, $$b=+\infty$$, and $$f(x)=x^{-s}$$, we have

$$\zeta(s)=\frac12+{1\over s-1}-s\int_1^\infty\overline B_1(x)x^{-s-1}\mathrm dx.$$

Therefore, we obtain

$$S_2=-2-\frac12-\zeta\left(\frac12\right)+O\left(1\over\sqrt n\right).\tag4$$

Now, plug (3) and (4) into (2), so we obtain

$$\sum_{k=1}^n\sqrt{\frac1k-\frac1n}=\frac\pi2\sqrt n+\zeta\left(\frac12\right)+O\left(1\over\sqrt n\right).$$

• Thank you very much for your original solution! I also tried to use the Euler-Maclaurin' formula but failed to make it rigorously and to get the answer Dec 15, 2022 at 8:12

This supplements the answer by @Gary, to justify the asymptotics conjectured there: $$S(n)\underset{n\to\infty}{\asymp}\frac\pi2+\sum_{k=0}^{(\infty)}\binom{k-1/2}{k}\frac{\zeta(1/2-k)}{n^{1/2+k}}.$$

From the answer, we take $$2\sum_{n=2}^\infty S_n z^n=f(z):=\operatorname{Li}_{1/2}^2(z)$$, hence $$S(n)=\frac1{4\pi i}\int_L\frac{f(z)}{z^{n+1}}\,dz$$, where $$L$$ is any simple contour encircling $$z=0$$ (in the domain of analyticity of the integrand).

In our case, $$f(z)$$ is analytic on $$\mathbb{C}\setminus\mathbb{R}_{\geqslant1}$$, and we have $$f(z)=O(\log|z|)$$ as $$z\to\infty$$. So, if we take $$L$$ to be the circle $$|z|=R$$ with a notch around $$\mathbb{R}_{\geqslant1}$$, the integral along the circle vanishes as $$R\to\infty$$ (assuming $$n>0$$). Thus, we can replace $$L$$ by $$L_1$$, where $$L_a$$ is the Hankel contour encircling $$\mathbb{R}_{\geqslant a}$$: $$S(n)=\frac1{4\pi i}\int_{L_1}\frac{f(z)}{z^{n+1}}\,dz\underset{z=e^w}{=}\frac1{4\pi i}\int_{L_0}e^{-nw}f(e^w)\,dw.$$

Now use $$f(e^w)=\big[(-\pi/w)^{-1/2}+\sum_{k=0}^\infty\zeta(1/2-k)w^k/k!\big]^2$$ for $$|w|<2\pi$$, noted in the answer. We replace $$L_0$$ by the contour consisting of the circle $$|w|=r$$ and the edges of the cut along $$[r,\infty)$$. As $$r\to0$$, the integral along the circle tends to $$(-2\pi i)\cdot(-\pi)$$, hence $$S(n)=\frac\pi2+\int_0^\infty e^{-nx}g(x)\,dx,\qquad g(x):=\frac1{4\pi i}\lim_{t\to 0^+}\big[f(e^{x+it})-f(e^{x-it})\big].$$

From the above we get $$g(x)=\pi^{-1/2}\sum_{k=0}^\infty\zeta(1/2-k)x^{k-1/2}/k!$$ for $$0, and the claimed asymptotics follows by Watson's lemma.

• Thank you very much for proving the proof of full form of asymptotics! Really, very impressive, and far beyond my skills ) Dec 15, 2022 at 22:10

This does not answer the question.

Let $$f(n)=\zeta \left(\frac{1}{2}\right)+\frac{\pi}{2}\sqrt n-\sum_{k=1}^n\sqrt{\frac{1}{k}-\frac{1}{n}}$$ Computing $$\Delta_p=10^{p+1}\,f\left(10^p\right)$$ the results $$\left( \begin{array}{cc} p & \Delta_p \\ 1 & 1.04871059298 \\ 2 & 1.04038414635 \\ 3 & 1.03952666777 \\ 4 & 1.03944068157 \\ 5 & 1.03943208058 \\ 6 & 1.03943122046 \\ 7 & 1.03943113444 \\ \end{array} \right)$$

• Your evaluation provides strongly evidence that $\zeta(1/2)$ is the limit Dec 14, 2022 at 9:16
• @Svyatoslav. For sure but no proof. What is amazing is the pattern. Fortunately, Gary provided something really interesting. I would like to know more. Cheers :-) Dec 14, 2022 at 9:22