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

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.
 A: 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 $L=2-I\left(\frac12\right)=-\zeta\left(\frac12\right)$.
A: 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 $L=1+\sum_{k\geq 1} v_k$. 
