Find $\lim_{n\to \infty}({1\over \sqrt{n^2+1}}+{1\over \sqrt{n^2+2}}+\cdots+{1\over \sqrt{n^2+n}})$ [closed]

Note

$$\lim_\limits{n\to \infty}\frac{n}{\sqrt{n^2+n}}\le\lim_\limits{n\to \infty}\left({1\over \sqrt{n^2+1}}+{1\over \sqrt{n^2+2}}+\cdots+{1\over \sqrt{n^2+n}}\right)\le\lim_\limits{n\to \infty}\frac{n}{\sqrt{n^2+1}}$$

Since $$\lim_\limits{n\to \infty}\frac{n}{\sqrt{n^2+n}}=\lim_\limits{n\to \infty}\frac{1}{\sqrt{1+\frac{1}{n}}}=1$$ and $$\lim_\limits{n\to \infty}\frac{n}{\sqrt{n^2+1}}=\lim_\limits{n\to \infty}\frac{1}{\sqrt{1+\frac{1}{n^2}}}=1$$

we have that the limit of the original is $1$ by the sandwich rule.

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[10px,#ffd]{\ds{\lim_{n \to \infty}\sum_{k = 1}^{n}{1 \over \root{n^{2} + k}}}} = \lim_{n \to \infty}\sum_{k = 1 + n^{2}}^{n + n^{2}}{1 \over \root{k}} = \lim_{n \to \infty}\pars{\sum_{k = 1}^{n + n^{2}}{1 \over \root{k}} - \sum_{k = 1}^{n^{2}}{1 \over \root{k}}} \\[5mm] = &\ \lim_{n \to \infty}\braces{\!\!\bracks{2\root{n + n^{2}} + \zeta\pars{1 \over 2} + {1 \over 2}\int_{n + n^{2}}^{\infty}\!\!{\braces{x} \over x^{3/2}}\,\dd x} - \bracks{2\root{n^{2}} + \zeta\pars{1 \over 2} + {1 \over 2}\int_{n^{2}}^{\infty}\!\!{\braces{x} \over x^{3/2}}\,\dd x}\!\!} \\[5mm] = &\ 2\lim_{n \to \infty}\pars{\root{n + n^{2}} - n} = 2\lim_{n \to \infty}{n \over \root{n + n^{2}} + n} = 2\lim_{n \to \infty}{1 \over \root{1/n + 1} + 1} = \bbx{1} \end{align}

Second Line: See Riemann Zeta Function Identity.

$$ \mbox{Note that}\quad 0 < \verts{\int_{N}^{\infty}{\braces{x} \over x^{3/2}}\,\dd x} < \int_{N}^{\infty}{\dd x \over x^{3/2}} = {2 \over \root{N}} \,\,\,\stackrel{\mrm{as}\ N\ \to\ \infty}{\large \to}\,\,\, {\large 0} $$