Prove that limit $\lim_{n\to\infty}\sqrt{4+\frac{1}{n^2}}+\sqrt{4+\frac{2}{n^2}}+\cdots+ \sqrt{4+\frac{n}{n^2}}-2n=\frac{1}{8}$ 
Let $$a_{n}=\sqrt{4+\dfrac{1}{n^2}}+\sqrt{4+\dfrac{2}{n^2}}+\cdots+
\sqrt{4+\dfrac{n}{n^2}}-2n,$$
  show that 
  $$\lim_{n\to\infty}a_{n}=\dfrac{1}{8}$$

My attempt:
Since
$$\sqrt{4+\dfrac{i}{n^2}}-2=\dfrac{\dfrac{i}{n^2}}{\sqrt{4+\dfrac{i}{n^2}}+2}=\dfrac{1}{\sqrt{4n^2+i}+2}\dfrac{i}{n}$$
then I can't work it. Thank you.

Let $\displaystyle \begin{array}{ccccc}
f & : & \mathbb R_+ & \to & \mathbb R_+ \\
 & & x & \mapsto & \sqrt{1+\frac{x}4 }-1\\
\end{array}$
Giving $\displaystyle a_n=2\sum_1^nf(\frac{k}{n^2})$
By Taylor expansion, there is a continuous function $\epsilon$,
such that $\epsilon(0)=0\;\;$ and $\; \;\displaystyle \forall x \in \mathbb R_+, f(x)=1+\frac{x}{8}+x\epsilon(x)-1$
Thus, $\displaystyle a_n=\frac14\sum_1^n\frac{k}{n^2}+2\sum_1^n\frac{k}{n^2}\epsilon(\frac{k}{n^2})$
And finally $$\displaystyle a_n=\frac18\frac{n(n+1)}{n^2}+2\sum_1^n\frac{k}{n^2}\epsilon(\frac{k}{n^2})$$

It remains to prove that $\displaystyle \sum_1^n\frac{k}{n^2}\epsilon(\frac{k}{n^2}) \to 0$
Let $\xi >0$
By continuity of $\epsilon$, there exists some $N\in \mathbb N$ such that $n\geq N \implies |\epsilon(\frac 1n)| \leq \xi$ 
Let $n\geq N$
Then, $\displaystyle \left |\sum_1^n\frac{k}{n^2}\epsilon(\frac{k}{n^2})\right| \leq \frac{\xi (n+1)}{2n}\leq \xi$

Therefore $$\displaystyle a_n=\frac18\frac{n(n+1)}{n^2} + o(1) = \frac{1}{8} + o(1)$$
