$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
\newcommand{\dd}{\mathrm{d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
\newcommand{\ic}{\mathrm{i}}
\newcommand{\mc}[1]{\mathcal{#1}}
\newcommand{\mrm}[1]{\mathrm{#1}}
\newcommand{\pars}[1]{\left(\,{#1}\,\right)}
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
\newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
\newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
\newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
I'll use the following identity ( from a previous @Vorhang user post ):
$$
\sum_{1 \leq\ d\ \leq\ x}{1 \over d^{s}} = {x^{1 - s} \over 1 - s} + \zeta\pars{s} -
{\braces{x} \over x^{s}} + s\int_{x}^{\infty}{\braces{u} \over u^{s + 1}}
\,\dd u\,,\qquad
\Re\pars{s} > 0.
$$
$$
\mbox{It leads to}\quad\bbox[#ffd,15px]{
\sum_{d = 1}^{n}{1 \over \root{d}} - 2\root{n} =
\zeta\pars{1 \over 2} +
{1 \over 2}\int_{n}^{\infty}{\braces{u} \over u^{3/2}}\,\dd u}
$$
$$
\mbox{However,}\quad
0 < \verts{{1 \over 2}\int_{n}^{\infty}{\braces{u} \over u^{3/2}}\,\dd u} <
{1 \over 2}\int_{n}^{\infty}{\dd u \over u^{3/2}} = {1 \over \root{n}}
\implies
\lim_{n \to \infty}
\bracks{{1 \over 2}\int_{n}^{\infty}{\braces{u} \over u^{3/2}}\,\dd u} = 0
$$
$$
\mbox{Then,}\quad
\bbx{\ds{%
\lim_{n \to \infty}\pars{\sum_{d = 1}^{n}{1 \over \root{d}} - 2\root{n}} =
\zeta\pars{1 \over 2}}} \approx -1.4604
$$