$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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Another one is given by
a well known identity:
\begin{align}
\sum_{k = 1}^{n}{1 \over k^{1/2}} & =
2\root{n} + \zeta\pars{1 \over 2} +
{1 \over 2}\int_{n}^{\infty}{\braces{x} \over x^{3/2}}\,\dd x
\end{align}
Note that
$\ds{0 <
{1 \over 2}\int_{n}^{\infty}{\braces{x} \over x^{3/2}}\,\dd x <
{1 \over 2}\int_{n}^{\infty}{\dd x \over x^{3/2}} = {1 \over \root{n}}}$ such that
$$
2\root{n} + \zeta\pars{1 \over 2} <
\sum_{k = 1}^{n}{1 \over k^{1/2}} <
2\root{n} + {1 \over \root{n}} + \zeta\pars{1 \over 2}
$$
and
$$
\sum_{k = 1}^{n}{1 \over k^{1/2}} \approx
2\root{n} + {1 \over 2\root{n}} + \zeta\pars{1 \over 2}\quad
\mbox{with an}\ absolute\ error\ < {1 \over 2\root{n}}
$$