Mathematical Reflections 2017, Issue 3 - Problem U409

Problem:
Evaluate $$\lim_{x \to 0} \dfrac{2\sqrt{1+x}+2\sqrt{2^2+x}+\ldots+2\sqrt{n^2+x}-n(n+1)}{x}.$$

Proposed by Nguyen Viet Hung, Hanoi University of Science, Vietnam

Solution:
Observe that
$$\renewcommand{\arraystretch}{2} \begin{array}{rcl} \sqrt{1+x}&=&1+\dfrac{1}{2}\cdot x+o(x) \\ \sqrt{2^2+x}=2\sqrt{1+\dfrac{x}{2^2}}&=&2+\dfrac{1}{2}\cdot\dfrac{x}{2}+o(x) \\ \vdots & \vdots & \vdots \\ \sqrt{n^2+x}=n\sqrt{1+\dfrac{x}{n^2}}&=& n+\dfrac{1}{2}\cdot\dfrac{x}{n}+o(x). \end{array}$$
So, $$\begin{array}{lll} 2\sqrt{1+x}+2\sqrt{2^2+x}+\ldots+2\sqrt{n^2+x}&=&2(1+2+\ldots+n)+x\left(1+\dfrac{1}{2}+\ldots+\dfrac{1}{n}\right)+o(x)\\&=&n(n+1)+x\left(1+\dfrac{1}{2}+\ldots+\dfrac{1}{n}\right)+o(x), \end{array}$$ which gives
$$\lim_{x \to 0} \dfrac{2\sqrt{1+x}+2\sqrt{2^2+x}+\ldots+2\sqrt{n^2+x}-n(n+1)}{x}=1+\dfrac{1}{2}+\ldots+\dfrac{1}{n}.$$