<s>[INST] 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
 [/INST] 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,
$$\lim_{n\to\infty} \sum_{k=n}^{9n-1}\frac{1}{\sqrt{k^2+1}+k}
= \lim_{n\to\infty} \frac12\left(\log(9n)-\log(n)\right) = \log 3.$$
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