Question: I have difficulty to understand the reason of following equality.

Can any one show me the reason of last equality ? 

The thing that ı did not understand is that how did they write as power of n because x are different than each other?

Picture is from Sheldon M. Ross stochastic process page 22.

[![enter image description here](https://i.stack.imgur.com/fSfWl.png)](https://i.stack.imgur.com/fSfWl.png)

Answer: 
See that

$${n\over\sqrt{n^2+2n}}\le \frac{1}{\sqrt{n²+2}}+\frac{1}{\sqrt{n²+4}}+\frac{1}{\sqrt{n²+6}} + \dots + \frac{1}{\sqrt{n²+2n}} \le {n\over\sqrt{n^2+2}}$$

And the limits of both ${n\over\sqrt{n^2+2n}}$ and ${n\over\sqrt{n^2+2}}$ is $1$. 

Therefore your limit is $1$.
$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,}
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 \newcommand{\ds}[1]{\displaystyle{#1}}
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 \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}\,}\,}
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\begin{align}
\lim\_{n \to \infty}\sum\_{k = 1}^{n}{1 \over \root{n^{2} + 2k}} & =
\lim\_{n \to \infty}\sum\_{k = 0}^{n - 1}{1 \over \root{n^{2} + 2k + 2}} =
{1 \over \root{2}}\,\lim\_{n \to \infty}
\sum\_{k = 0}^{n - 1}{1 \over \bracks{k + \pars{n^{2}+ 2}/2}^{1/2}}
\\[5mm] & =
{1 \over \root{2}}\,\lim\_{n \to \infty}
\bracks{\Phi\pars{1^{-},{1 \over 2},{n^{2} + 2 \over 2}} -
\Phi\pars{1^{-},{1 \over 2},{n^{2} + 2 \over 2} + n}}
\end{align}
where we used a [Lerch Zeta Function $\ds{\Phi}$ identity](https://en.wikipedia.org/wiki/Lerch_zeta_function#Identities).
$\ds{\Phi\pars{1^{-},s,a}}$ satisfies the
[asymptotic expression](https://en.wikipedia.org/wiki/Lerch_zeta_function#Integral_representations):
$$
\Phi\pars{1^{-},s,a} \sim {1 \over \pars{s - 1}\, a^{s - 1}}\quad\mbox{as}\quad a \to \infty 
$$

---

Then,
\begin{align}
\lim\_{n \to \infty}\sum\_{k = 1}^{n}{1 \over \root{n^{2} + 2k}} & =
{1 \over \root{2}}\,\lim\_{n \to \infty}\pars{%
-2\root{n^{2} + 2 \over 2} + 2\root{{n^{2} + 2 \over 2} + n}}
\\[5mm] & =
\lim\_{n \to \infty}\pars{\root{n^{2} + 2n + 2} - \root{n^{2} + 2}}
\\[5mm] & =
\lim\_{n \to \infty}{2 \over \root{1 + 2/n + 2/n^{2}} + \root{1 + 2/n^{2}}} =
\bbx{\ds{1}}
\end{align}