Find $\lim_{n\to \infty}({1\over \sqrt{n^2+1}}+{1\over \sqrt{n^2+2}}+\cdots+{1\over \sqrt{n^2+n}})$ strong textFind $\lim_\limits{n\to \infty}\left({1\over \sqrt{n^2+1}}+{1\over \sqrt{n^2+2}}+\cdots+{1\over \sqrt{n^2+n}}\right)$.
I do know it is bounded by $1$. I tried using the sandwich rule with no success. How can I solve it?
 A: 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.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\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.

$$
\mbox{Note that}\quad
0 < \verts{\int_{N}^{\infty}{\braces{x} \over x^{3/2}}\,\dd x} <
\int_{N}^{\infty}{\dd x \over x^{3/2}} = {2 \over \root{N}}
\,\,\,\stackrel{\mrm{as}\ N\ \to\ \infty}{\large \to}\,\,\, {\large 0}
$$
A: Note that
$$\left|{1\over\sqrt{n^2+k}}-{1\over n}\right|=
\left|{n-\sqrt{n^2+k}\over n\sqrt{n^2+k}}\right|
={k\over n\sqrt{n^2+k}(n+\sqrt{n^2+k})}\lt{k\over2n^3}$$
Therefore
$$\left|\sum_{k=1}^n{1\over\sqrt{n^2+k}}-1 \right|
=\left|\sum_{k=1}^n\left({1\over\sqrt{n^2+k}}-{1\over n}\right) \right|
\le\sum_{k=1}^n\left|{1\over\sqrt{n^2+k}}-{1\over n}\right|
\lt\sum_{k=1}^n{k\over2n^3}={n(n+1)\over4n^3}\to0$$
as $n\to\infty$. It follows from the definition of limit that
$$\lim_{n\to\infty}\sum_{k=1}^n{k\over\sqrt{n^2+k}}=1$$
