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.
 A: Let us make the problem more general and search the limit of $$a_n=\sum_{i=1}^{i=n}\sqrt {k^2+\frac{i}{n^2}} -k~n$$ In the same spirit as Vladimir's answer, for sufficiently large values of $n$, we can write $$\sqrt {k^2+\frac{i}{n^2}}=k+\frac{i}{2 k n^2}+O\left(\left(\frac{1}{n}\right)^4\right)$$ So $$a_n=\sum_{i=1}^{i=n}k+\sum_{i=1}^{i=n}\frac{i}{2kn^2}-kn=\frac{1}{2kn^2}\sum_{i=1}^{i=n}i=\frac{n(n+1)}{4kn^2}$$ and then $$\lim_{n\to\infty}a_{n}=\dfrac{1}{4k}$$ By pure luck, I found that there is a closed form expression for $a_n$; this involves the Hurwitz zeta function.
A: 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.
A: 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 $$\displaystyle a_n=\frac18\frac{n(n+1)}{n^2} + o(1) = \frac{1}{8} + o(1)$$
A: $\sqrt{4+\frac k{n^2}}-2=2(1+\frac k{8n^2}+o(\frac k{8n^2})-1)=\frac k{4n^2}(1+o(1))$
(use the formula $\sqrt{1+x}=1+\frac x2+\dots$)
A: Distribute $2n$ amongst the terms. 
A general term would look like $$\sqrt{4+\frac{x}{n^2}}-\sqrt{4}=\frac{\frac{x}{n^2}}{\sqrt{4+\frac{x}{n^2}}+\sqrt{4}}$$
Now $$\frac{\frac{x}{n^2}}{\sqrt{4+\frac{n}{n^2}}+\sqrt{4}}<\frac{\frac{x}{n^2}}{\sqrt{4+\frac{x}{n^2}}+\sqrt{4}}<\frac{\frac{x}{n^2}}{\sqrt{4}+\sqrt{4}}$$
Using the sandwich theorem, it is easy to see that the limit if $\frac{1}{8}$.
A: Let $0 < \alpha < \frac{1}{4}$ then for $n$ large enough and $k \in \{1,\ldots,n\}$ $$2 + \frac{\alpha \, k}{n^2} < \sqrt{4+\frac{k}{n^2}} < 2 + \frac{k}{4 n^2}.$$ The limit  now follows from $$1+2+\ldots+n = \frac{n(n+1)}{2}$$ and taking $\alpha \uparrow \frac{1}{4}$.
A: 
My solution: Note 
  $$\sqrt{4+\dfrac{k}{n^2}}<\sqrt{2^2+\dfrac{k}{n^2}+(\dfrac{k}{4n^2})^2}=2+\dfrac{k}{4n^2}$$
  $$\sqrt{4+\dfrac{k}{n^2}}=\sqrt{4+\dfrac{k-1}{n^2}+\dfrac{1}{n^2}}>\sqrt{2^2+\dfrac{k-1}{n^2}
+(\dfrac{k-1}{4n^2})^2}=2+\dfrac{k-1}{4n^2}$$
  and
  $$\lim_{n\to\infty}\dfrac{k-1}{4n^2}=\lim_{n\to\infty}\dfrac{k}{4n^2}=\dfrac{1}{8}$$

