$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ Comprobemos lo que dice Euler-Maclaurin: \begin{align} &\color{#0000ff}{\large\sum_{k = 1}^{n}{1 \over \root{n^{2} + n - k^{2}}}}={1 \over \root{n}} + \sum_{k = 1}^{n - 1}{1 \over \root{n^{2} + n - k^{2}}} \\[3mm]&= {1 \over \root{n}} + \overbrace{\int_{0}^{n}{\dd k \over \root{n^{2} + n - k^{2}}}}^{\ds{\arcsin\pars{n \over\root{n^{2} + n}}}} - \half\pars{{1 \over \root{n^{2} + n}} + {1 \over \root{n}}} + {1 \over 12}\,{1 \over \root{n}}\\[3mm]& - {1 \over 720}\pars{{15 \over \root{n}} + {9 \over n\root{n}}} + \cdots \color{#0000ff}{\large\stackrel{n \to \infty}{\to} {\pi \over 2}} \end{align}
Parece que $\pi/2$ es el resultado correcto pero la convergencia, en la parte numérica, es muy lenta. Incluso con $n = 10^{8}$ la diferencia respecto a $\pi/2$ es $\approx -0.00135758$ .