How prove this limit $\lim_{n\to\infty}\sum_{k=n}^{9n-1}\frac{1}{\sqrt{k^2+1}+k}=\ln{3}$ 
show  that: this limit
  $$\lim_{n\to\infty}\sum_{k=n}^{9n-1}\dfrac{1}{\sqrt{k^2+1}+k}=\ln{3}$$

my idea: since
$$\dfrac{1}{\sqrt{k^2+1}+k}=\sqrt{k^2+1}-k$$
then I can't.Thank you
 A: 
Observation. If $a_{k} \sim c/k$ as $k \to \infty$, then we have
  $$ \lim_{n\to\infty} \sum_{pn < k \leq qn} a_{k} = c\log(q/p). $$

Proof. Given $\epsilon > 0$, choose $N$ large so that $k > pN$ then
$$ \frac{c-\epsilon}{k} \leq a_{k} \leq \frac{c+\epsilon}{k}. \tag{1} $$
Now define $H(x) = \sum_{k \leq x} \frac{1}{k}$ and recall that $H(x) = \gamma + \log x + o(1)$ as $x \to \infty$. Summing (1) over $pn < k \leq qn$ with $n \geq N$, we have
$$ (c-\epsilon)\{ H(qn) - H(pn) \} \leq \sum_{pn < k \leq qn} a_{k} \leq (c+\epsilon)\{ H(qn) - H(pn) \}. $$
Taking $n\to\infty$, it follows that
$$ (c-\epsilon) \log(q/p)
\leq \liminf_{n\to\infty} \sum_{pn < k \leq qn} a_{k}
\leq \limsup_{n\to\infty} \sum_{pn < k \leq qn} a_{k}
\leq (c+\epsilon) \log(q/p). $$
Letting $\epsilon \to 0$, the conclusion follows. ////
Now you can apply this observation to your problem.
A: Notice $\frac{1}{\sqrt{k^2+1}+k}$ is monotonic decreasing, we have
$$\sum_{k=n}^{9n-1}\frac{1}{\sqrt{k^2+1}+k} \ge
\int_n^{9n}\frac{dx}{\sqrt{x^2+1}+x} \ge \sum_{k=n+1}^{9n}\frac{1}{\sqrt{k^2+1}+k}\\
$$
This implies
$$\left| \sum_{k=n}^{9n-1}\frac{1}{\sqrt{k^2+1}+k} - 
\int_n^{9n}\frac{dx}{\sqrt{x^2+1}+x} \right|
\le \frac{1}{\sqrt{n^2+1}+n} - \frac{1}{\sqrt{(9n)^2+1}+9n} < \frac{1}{2n}
$$
and hence 
$$\lim_{n\to\infty} \sum_{k=n}^{9n-1}\frac{1}{\sqrt{k^2+1}+k}
= \lim_{n\to\infty} \int_n^{9n}\frac{dx}{\sqrt{x^2+1}+x}\tag{*1}$$
Notice in the indefinite integral,
$$\int \frac{dx}{\sqrt{x^2+1}+x} = \int\left(\sqrt{x^2+1}-x\right) dx
= \frac12\left( x(\sqrt{x^2+1} -x) + \log(\sqrt{x^2+1}+x) \right)
= \frac12\left[ \color{blue}{\frac{x}{\sqrt{x^2+1}+x} + \log(\sqrt{1+x^{-2}}+1)} + \log x\right]
$$
The first two term (in blue) converges to a constant $\frac12 + \log 2$ as $x\to\infty$.
they won't contribute to the limit in RHS of $(*1)$. As a result,
$$\lim_{n\to\infty} \sum_{k=n}^{9n-1}\frac{1}{\sqrt{k^2+1}+k}
= \lim_{n\to\infty} \frac12\left(\log(9n)-\log(n)\right) = \log 3.$$
A: By multiplying and dividing the sum by n, we notice that this is nothing else but the Riemann sum for $~\displaystyle\int_1^9\frac{dx}{\sqrt{x^2}+x}=\int_1^9\frac{dx}{2x}=\bigg[\frac{\ln x}2\bigg]_1^9=\ln3$.
A: Let $a_n$ denote the $n$-th term. Then this limit follows from the inequalities
$$
\int_n^{9n}\frac{dx}{2x+1}\leq\sum_{k=n}^{9n-1}\frac{1}{2k+1}\leq a_n \leq\sum_{k=n}^{9n-1}\frac{1}{2k}\leq\int_{n-1}^{9n-1}\frac{dx}{2x}.
$$
A: $\newcommand{\+}{^{\dagger}}
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$\ds{\lim_{n\to\infty}\sum_{k=n}^{9n - 1}{1 \over \root{k^{2} + 1} + k}
     =\ln\pars{3}:\ {\large ?}}$

\begin{align}
\mbox{With}\quad S_{N} \equiv \sum_{k = 1}^{N}{1 \over \root{k^{2} + 1} + k}\,,
\qquad
\lim_{n\to\infty}\sum_{k=n}^{9n - 1}{1 \over \root{k^{2} + 1} + k}=
\lim_{n \to \infty}\pars{S_{9n - 1} - S_{n - 1}}
\end{align}

\begin{align}
S_{N}&=\sum_{k = 1}^{N}\pars{{1 \over \root{k^{2} + 1} + k} - {1 \over 2k}}
+\half\,H_{N}
\\[3mm]& H_{N}:\mbox{Harmonic Number} = \Psi\pars{N + 1} + \gamma
\end{align}
$\ds{\Psi\pars{z}\mbox{: Digamma Function.}\quad
     \gamma\mbox{: Euler-Mascheroni Constant}}$.

$\ds{\Psi\pars{z} \sim\ln\pars{z}\ \mbox{when}\ \verts{z} \gg 1}$ and
  $\ds{\pars{S_{N} - \half\,H_{N}}}$ converges in the limit $\ds{N \to \infty}$.
  Then,
  \begin{align}&\color{#66f}{\large%
\lim_{n\to\infty}\sum_{k=n}^{9n - 1}{1 \over \root{k^{2} + 1} + k}}
=\half\,\lim_{n \to \infty}\bracks{\Psi\pars{9n} - \Psi\pars{n}}
=\half\,\lim_{n \to \infty}\bracks{\ln\pars{9n} - \ln\pars{n}}
\\[3mm]&=\color{#66f}{\large\ln\pars{3}}
\end{align}

