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.

From$\let\leq\leqslant\let\geq\geqslant$ HM-GM-AM we have
$$\frac x{\frac x4+1}\leq\sqrt x\leq\frac x4+1\qquad(x>0)$$
with equality iff $x=4$.
Using this (the intuition is that as the arguments of our square roots get closer to $4$, these bounds will be very close), we get
$$\sum\frac{4+\frac k{n^2}}{2+\frac k{4n^2}}\leq a_n+2n\leq\sum\left(2+\frac k{4n^2}\right).$$
The right-most expression can be evaluated as $2n+\frac18\frac{n(n+1)}{n^2}$. The left-most sum can be estimated further using
$$\sum\frac{4+\frac k{n^2}}{2+\frac k{4n^2}}=\sum\left(2+2\frac k{8n^2+k}\right)\geq\sum\left(2+2\frac k{8n^2+n}\right)=2n+\frac{n(n+1)}{8n^2+n}.$$
In summary,
$$\frac{n(n+1)}{8n^2+n}\leq a_n\leq\frac18\frac{n(n+1)}{n^2}.$$
The conclusion follows.

Note that this is essentially the same as considering the Taylor expansion of $\sqrt x$ around $x=4$. In fact the HM-GM-AM inequalities give two functions, $\frac x{\frac x4+1}$ and $\frac x4+1$, with the same Taylor expansion up to terms of first order. This can be seen, for example, by rewriting the difference between two such functions as a rational function in $\sqrt x$ with $4$ as root with multiplicity $2$: $\frac x4+1-\sqrt x=\frac14(\sqrt x-\sqrt4)^2$ and $\sqrt x-\frac x{\frac x4+1}=\frac{\sqrt x}{x+4}(\sqrt x-\sqrt4)^2$.
This means that HM-GM-AM is a good approximation, since the limit to be found requires only constants and first order terms. However if the question suggested a second order approximation, this kind of ad-hoc methods with AM-GM-like inequalities become far more complicated, and Taylor approximation would be the way to go.
