Find two constant$c_1, c_2$ to make the inequality hold true for all N Show that the following inequality holds for all integers $N\geq 1$
$\left|\sum_{n=1}^N\frac{1}{\sqrt{n}}-2\sqrt{N}-c_1\right|\leq\frac{c_2}{\sqrt{N}}$
where $c_1,c_2$ are some constants.
It's obvious that by dividing $\sqrt{N}$, we can get $\left|\frac{1}{N}\sum_{n=1}^N \frac{1}{\sqrt{n/N}} -2 - \frac{c_1}{\sqrt{N}}\right|\leq \frac{c_2}{N}$ the first 2 terms in LHS is a Riemann Sum of $\int_{0}^1\frac{1}{\sqrt{x}}dx$. But what is the next step?
I have not seen a complete answer of this problem in this site or google, so please give me some hints.
 A: As Hans Engler commented, the problem is not entirely elementary.
Let us check the pieces of the expression $$\frac 1 N \sum_{n=1}^n \frac 1 {\sqrt {\frac n N}}=\frac 1 {\sqrt { N}}\sum_{n=1}^n \frac 1 {\sqrt { n}}=\frac{1}{\sqrt{N}}H_N^{\left(\frac{1}{2}\right)}$$ where appear the generalized harmonic numbers. So, the lhs write $$\frac{-c_1+H_N^{\left(\frac{1}{2}\right)}-2 \sqrt{N}}{\sqrt{N}}$$ For large values of $N$, using the asymptotics of harmonic numbers (this is an altenating series ),  this can write $$ \left(\zeta \left(\frac{1}{2}\right)-c_1\right)\sqrt{\frac{1}{N}}+\frac{1}{2
   N}-\frac{1}{24 N^2}+O\left(\frac{1}{N^{5/2}}\right)$$ Compared to the rhs, just as Hans Engler commented, this seems to lead to $$c_1=\zeta \left(\frac{1}{2}\right)\qquad , \qquad c_2=\frac 12$$ Checking for small values of $N$, this seems to work properly for all $N$.
A: Restating the problem in equivalent terms, we have to find the asymptotic behaviour of $H^{(1/2)}_N$.
For such a task, we may apply creative telescoping. Since the primitive of $\frac{1}{\sqrt{x}}$ is given by $2\sqrt{x}$, we may expect to be able to write $\frac{1}{\sqrt{n}}$ as $2\sqrt{n}-2\sqrt{n-1}$ (that is a telescopic term) plus a small error term. Indeed we have:
$$ \frac{1}{\sqrt{n}} = \left(2\sqrt{n}-2\sqrt{n-1}\right)-\frac{1}{\sqrt{n}\left(\sqrt{n}+\sqrt{n-1}\right)^2} \tag{1}$$
hence:
$$ H^{(1/2)}_N = \sum_{n=1}^{N}\frac{1}{\sqrt{n}} = 2\sqrt{N}-\sum_{n=1}^{N}\frac{1}{\sqrt{n}(\sqrt{n}+\sqrt{n-1})^2} \tag{2}$$
but the last series is clearly converging as $N\to +\infty$, so by setting
$$ c_1 = -\sum_{n=1}^{+\infty}\frac{1}{\sqrt{n}(\sqrt{n}+\sqrt{n-1})^2}\tag{3} $$
we may write:
$$ H^{(1/2)}_N = 2\sqrt{N}+c_1+\color{red}{\sum_{n>N}\frac{1}{\sqrt{n}(\sqrt{n}+\sqrt{n-1})^2}}\tag{4}$$
and it is enough to prove that the red series is bounded above by $\frac{c_2}{\sqrt{N}}$. That can be easily achieved by comparing the red series with an integral, or by creative telescoping again.
So, if we do not have to find the exact values of $c_1$ and $c_2$, the question is very simple to tackle through H. Wilf's favourite technique. If we exploit Abel's summation formula, with a few extra work, we may also find 
$$ c_1=\zeta\left(\frac{1}{2}\right)=(1+\sqrt{2})\sum_{n\geq 1}\frac{(-1)^n}{\sqrt{n}},\qquad c_2=\frac{1}{2} \tag{4} $$
as already found by monsieur Leibovici.
