TITLE: How prove this limit $\lim_{n\to\infty}\sum_{k=n}^{9n-1}\frac{1}{\sqrt{k^2+1}+k}=\ln{3}$
QUESTION [5 upvotes]: 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

REPLY [2 votes]: $\newcommand{\+}{^{\dagger}}
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$\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}