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.

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 $N$.
