Closed form of $\lim\limits_{n\to\infty}\left(\int_0^{n}\frac{{\rm d}k}{\sqrt{k}}-\sum_{k=1}^n\frac1{\sqrt k}\right)$

Question

Show that $$ L=\lim_{s\rightarrow\infty}\left(\int_0^s\frac{ds'}{\sqrt{s'}}-\sum_{s'=1}^s\frac{1}{\sqrt{s'}}\right) = 1.460\ldots $$

My attempts: To begin, rewriting the limit of the form $$ L=\lim_{\epsilon\rightarrow0}\left(\int_0^{\infty}\frac{e^{-\epsilon s'}}{\sqrt{s'}}ds'-\sum_{s'=1}^{\infty}\frac{e^{-\epsilon s'}}{\sqrt{s'}}\right) $$ where $$ \int_0^{\infty}\frac{e^{-\epsilon s'}}{\sqrt{s'}}ds' = \int_0^{\infty}\frac{e^{-\epsilon s^2}}{s}d(s^2) = \sqrt{\frac{\pi}{\epsilon}} $$ and $$ \sum_{s'=1}^{\infty}\frac{e^{-\epsilon s'}}{\sqrt{s'}} = \sum_{s'=1}^{\infty}{e^{-\epsilon s'}}\int_0^{\infty}e^{-\sqrt{s'}t}dt = \int_0^{\infty}\left(\sum_{s'=1}^{\infty}{e^{-\epsilon s'-\sqrt{s'}t}}\right)dt $$ Second, the identity $$ \sum_{s=1}^{\infty}\frac{(1-\epsilon)^s}{\sqrt{s}}=\sqrt{\frac{\pi}{\epsilon}}(1+O(\epsilon)); $$ may be of some help.

Thirdly, the limit is somehow $-\zeta(1/2)=1.46035\cdots$ where $\zeta(s)$ is the Riemann zeta function.

Answer 1

Let us represent the finite sum $\sum_{k=1}^n\frac{1}{\sqrt{k}}$ as \begin{align*}\sum_{k=1}^n\frac{1}{\sqrt{k}}&=\sum_{k=1}^n\frac{1}{\sqrt \pi}\int_0^{\infty} \frac{e^{-kx}dx}{\sqrt x}=\frac{1}{\sqrt \pi}\int_0^{\infty} e^{-x}\frac{1-e^{-(n+1)x}}{1-e^{-x}}\frac{dx}{\sqrt x}=\\ &=\frac{1}{\sqrt \pi}\int_0^{\infty} \frac{1-e^{-(n+1)x}}{\sqrt x}\left(\frac{1}{e^x-1}-\frac{e^{-x}}x+\frac{e^{-x}}x\right)dx=\\&= \frac{1}{\sqrt \pi}\int_0^{\infty} \frac{1-e^{-(n+1)x}}{\sqrt x}\left(\frac{1}{e^x-1}-\frac{e^{-x}}x\right)dx+\frac{1}{\sqrt \pi}\int_0^{\infty} \frac{e^{-x}-e^{-(n+2)x}}{x\sqrt x}dx=\\ &=\frac{1}{\sqrt \pi}\int_0^{\infty} \frac{1-e^{-(n+1)x}}{\sqrt x}\left(\frac{1}{e^x-1}-\frac{e^{-x}}x\right)dx+2\sqrt{n+2}-2. \end{align*} Since $\sqrt{n+2}-\sqrt n\to 0$ as $n\to \infty$, the limit we are looking for is given by $$L=2-\frac{1}{\sqrt \pi}\int_0^{\infty} \left(\frac{1}{e^x-1}-\frac{e^{-x}}x\right)\frac{dx}{\sqrt x}.$$ It is not difficult to relate this expression to zeta value $\zeta\left(\frac12\right)$. Indeed, the integral $$I(s)=\frac{1}{\sqrt \pi}\int_0^{\infty} x^{s-1}\left(\frac{1}{e^x-1}-\frac{e^{-x}}x\right)dx$$ converges and defines an analytic function of $s$ in the region $\Re s>0$. Furthermore for $\Re s>1$ we can break it into two separate pieces, which leads to evaluation \begin{align*}I(s)=\color{blue}{\frac{1}{\sqrt \pi}\int_0^{\infty} \frac{x^{s-1} dx}{e^x-1}}-&\color{red}{\frac{1}{\sqrt \pi}\int_0^{\infty}x^{s-2}e^{-x}dx}=\frac{\color{blue}{\Gamma(s)\zeta(s)}-\color{red}{\Gamma(s-1)}}{\sqrt\pi}=\\ &=\frac{\Gamma(s)}{\sqrt\pi}\left[\zeta(s)-\frac{1}{s-1}\right]. \end{align*} This finally gives $L=2-I\left(\frac12\right)=-\zeta\left(\frac12\right)$.

Answer 2

By the Euler-Maclaurin formula $$\sum_{k=1}^{n}\frac{1}{\sqrt k }-\int _{1}^n \frac{dt}{\sqrt t }=\frac{1}{2}+\frac{1}{2\sqrt n }-\int _1^n \frac{B_1(t-\lfloor t\rfloor)}{2t^{3/2}}\ dt$$

where $B_1$ is the first Bernoulli polynomial. Then, $$\sum_{k=1}^n \frac{1}{\sqrt k}-\int _0 ^n \frac{dt}{\sqrt t}=-\int _0 ^1 \frac{dt}{\sqrt t}+\frac{1}{2}+\frac{1}{2\sqrt n }-\int _1^n \frac{B_1(t-\lfloor t\rfloor )}{2t^{3/2}}\ dt$$ and upon integrating the RHS, we have

$$\sum_{k=1}^n \frac{1}{\sqrt k}-\int_0^n\frac{dt}{\sqrt t} =-\frac{3}{2} + \frac{1}{2\sqrt n }-\int _1^n \frac{B_1(t-\lfloor t\rfloor )}{2t^{3/2}}\ dt.$$

Next, applying the identity $\int_1^n=\int_1^{+\infty}-\int _n^{+\infty}$ to the integral on the right and noting that the periodic Bernoulli polynomial is a bounded function $$\int _n^{+\infty} \frac{B_1(t-\lfloor t\rfloor )}{2t^{3/2}}\ dt=O\left(\frac{1}{\sqrt n}\right)$$ as $n\to +\infty$. Thus, $$\sum_{k=1}^n \frac{1}{\sqrt k}-\int_0^n\frac{dt}{\sqrt t} =-\frac{3}{2} -\int _1^{+\infty} \frac{B_1(t-\lfloor t\rfloor )}{2t^{3/2}}\ dt +O\left(\frac{1}{\sqrt n}\right).$$ as $n\to +\infty$. Now employing the Euler-Maclaurin formula for $\zeta(s)$, $$\zeta\left(s\right) -\int_1^{+\infty}\frac{dt}{t^s}=\frac{1}{2}-s\int_1^{+\infty}\frac{B_1(t-\lfloor t\rfloor)}{t^{s+1}}\ dt$$ for $\text{Re} (s) >1$, or upon integrating and rearranging $$-s\int_1^{+\infty}\frac{B_1(t-\lfloor t\rfloor)}{t^{s+1}}\ dt=\zeta (s)-\frac{1}{s-1}-\frac{1}{2}$$ for $\text{Re} (s)>0$ and $s\neq1$, via analytic continuation. Substituting $s=1/2$, we obtain $$-\int_1^{+\infty}\frac{B_1(t-\lfloor t\rfloor)}{2t^{3/2}}\ dt=\zeta(1/2)+\frac{3}{2}.$$ Therefore, $$\sum_{k=1}^n \frac{1}{\sqrt k}-\int_0^n\frac{dt}{\sqrt t} = \zeta(1/2)+O\left(\frac{1}{\sqrt n}\right)$$ as $n\to +\infty$.

Answer 3

I suppose that you want to show that the sequence $u_n=2\sqrt{n}-\sum_{k=1}^n \frac{1}{\sqrt{k}}$ is convergent. We compute $v_n=u_{n+1}-u_n$. I have found $$v_n=\frac{1}{(\sqrt{n+1})(\sqrt{n}+\sqrt{n+1})^2}$$ As the series $\sum v_n$ is convergent, the sequence $u_n$ is convergent, and the limit $L$ is $L=1+\sum_{k\geq 1} v_k$.