Sequence: $u_n=\sum_{k=n}^{2n}\frac{k}{\sqrt{n^2+k^2}}$ 
Study the following sequence of numbers:
  $$\forall n\in\mathbb{N}, u_n=\sum_{k=n}^{2n}\frac{k}{\sqrt{n^2+k^2}}$$

I tried to calculate $u_{n+1}-u_n$, but I couldn't simplify the expression.
Plotting the sequence shows arithmetic (or seems to be an) progression.
 A: We have that
$$\sum_{k=n}^{2n}\frac{k}{\sqrt{n^2+k^2}} \ge n\cdot \frac{n}{\sqrt{2n^2}}=\frac{n}{\sqrt 2}$$
form which we conclude that $u_n \to \infty$, we can also obtain that
$$\frac{n}{\sqrt{2}}\le \sum_{k=n}^{2n}\frac{k}{\sqrt{n^2+k^2}} \le \frac{2n}{\sqrt{5}}$$
and by Riemann sum since
$$\lim_{n\to \infty }\frac1n \sum_{k=n}^{2n}\frac{k}{\sqrt{n^2+k^2}}= \lim_{n\to \infty }\frac1n\sum_{k=n}^{2n}\frac{\frac kn}{\sqrt{1+\frac{k^2}{n^2}}}=\int_1^2 \frac{x}{\sqrt{1+x^2}}dx=\sqrt 5 - \sqrt 2$$
we obtain
$$\sum_{k=n}^{2n}\frac{k}{\sqrt{n^2+k^2}}\sim (\sqrt 5 - \sqrt 2)n$$
A: $$\forall n\in\mathbb{N}, u_n=\sum_{k=n}^{2n}\frac{k}{\sqrt{n^2+k^2}}\\ u_{n+1}=\sum_{k=n+1}^{2n+2}\frac{k}{\sqrt{(n+1)^2+k^2}}\\u_{n+1}-u_{n}=\underbrace{\sum_{k=n+1}^{2n+2}\frac{k}{\sqrt{(n+1)^2+k^2}}}_{n+2 \space terms}-\underbrace{\sum_{k=n}^{2n}\frac{k}{\sqrt{n^2+k^2}}}_{n+1 \space  terms}\\$$
A: Using the Euler-Maclaurin formula we can get more detailed asymptotics, e.g.:
$$ u_n =  \left( \sqrt {5}-\sqrt {2} \right) n
+ \frac{\sqrt{5}}{5} + \frac{\sqrt{2}}{4}
+\left(\frac{\sqrt{5}}{300}- \frac{\sqrt{2}}{48}  \right) n^{-1} +
\left(-\frac{\sqrt{5}}{10000} + \frac{\sqrt{2}}{1280} \right) n^{-3}
+O \left( {n}^{
-5} \right) $$
