How to estimate $ \left(1 + \sqrt{2} + \sqrt{3} + \dots + \sqrt{n} \right) - \frac{2}{3} n \sqrt{n}$? How do I estimate the error for the sum of reciprocals of square roots.  From calculus we know that:
$$ \int_0^n \sqrt{x} \, dx  = \frac{2}{3} x\sqrt{x} \;\;\Bigg|_{x=0}^{x=n} = \frac{2}{3}n\sqrt{n}$$
I forget the name - midpoint rule, trapezoid rule ?? - basically we want to approximate the integral as a Riemann sum.  How do we estimate the error?
$$ \left(1 + \sqrt{2} + \sqrt{3} + \dots +  \sqrt{n} \right)  - \frac{2}{3} n \sqrt{n}$$
To give you a sense of how much we are losing on this approximation, let's   draw two pictures.


We are losing all the gray stuff in our approximation, which is quite a lot!  I don't really care about the integral, what's important is the difference between all the stuff we are adding and square root of $n$. 
The yellow triangle has base $1$ and height $\sqrt{14} - \sqrt{13}$ so the area is:
$$ A = \frac{1}{2} \times b \times h = \frac{1}{2} \times 1 \times (\sqrt{\color{#E0E070}{14}} - \sqrt{\color{#E0E070}{13}}) = \frac{1}{2}(\sqrt{n+1} - \sqrt{n}) \approx \frac{1}{4 \sqrt{n}}$$
This suggests the total of all errors is about $\propto \sqrt{\color{#D22}{n}}$ which is not a small amount.  Can anyone get the constant of proportionality?
The Euler-Maclaurin machine does crank out such error estimates, but the square root function is not so strange. Can we derive such an estimate in this specific case using basic standard inequalities?
 A: A trapezoidal rule gives a much better approximation: $$\sum_1^n i^{1/2}\approx \frac 1 2(f(1)+f(n))+\int_1^n f(x)\,dx= \frac 1 2 +\frac 1 2 \sqrt n +\frac 2 3 n^{3/2}-\frac 2 3$$ which recovers the biggest component of the error and ends up with an error bound by a small constant.
Error analysis can also improve the above estimate to
$$\sum_1^n i^{1/2}=\frac 2 3n^{3/2}+\frac 1 2n^{1/2}-\frac 1 6-\frac 1 {24}+\frac 1{24}n^{-1/2}+E_n$$
where non-negative $E_n$ is tiny ($<5\cdot 10^{-4}$).
A: Using the Binomial Theorem, we get
$$
\begin{align}
k^{3/2}-(k-1)^{3/2}
&=k^{3/2}\left[1-\left(1-\frac1k\right)^{3/2}\right]\\
&=k^{3/2}\left[\frac3{2k}-\frac3{8k^2}-\frac1{16k^3}+O\left(\frac1{k^4}\right)\right]\\
&=\frac32k^{1/2}-\frac38k^{-1/2}-\frac1{16}k^{-3/2}+O\left(k^{-5/2}\right)\tag{1}
\end{align}
$$
and
$$
\begin{align}
k^{1/2}-(k-1)^{1/2}
&=k^{1/2}\left[1-\left(1-\frac1k\right)^{1/2}\right]\\
&=k^{1/2}\left[\frac1{2k}+\frac1{8k^2}+O\left(\frac1{k^3}\right)\right]\\
&=\frac12k^{-1/2}+\frac18k^{-3/2}+O\left(k^{-5/2}\right)\tag{2}
\end{align}
$$
and
$$
\begin{align}
k^{-1/2}-(k-1)^{-1/2}
&=k^{-1/2}\left[1-\left(1-\frac1k\right)^{-1/2}\right]\\
&=k^{-1/2}\left[-\frac1{2k}+O\left(\frac1{k^2}\right)\right]\\
&=-\frac12k^{-3/2}+O\left(k^{-5/2}\right)\tag{3}
\end{align}
$$
Combining $(1)$, $(2)$, and $(3)$ gives
$$
\begin{align}
&\small\left(\frac23k^{3/2}+\frac12k^{1/2}+\frac1{24}k^{-1/2}\right)-\left(\frac23(k-1)^{3/2}+\frac12(k-1)^{1/2}+\frac1{24}(k-1)^{-1/2}\right)\\[6pt]
&\small=k^{1/2}+O\left(k^{-5/2}\right)\tag{4}
\end{align}
$$
Summing, we get that
$$
\sum_{k=1}^nk^{1/2}=\frac23n^{3/2}+\frac12n^{1/2}+C+\frac1{24}n^{-1/2}+O\left(n^{-3/2}\right)\tag{5}
$$
where $C=\zeta\left(-\frac12\right)=-0.207886224977354566\dots$

Why $\boldsymbol{\zeta\left(-\frac12\right)}$?
We can rewrite the reasoning given in this answer, which uses the Euler-Maclaurin Sum Formula, but use the Binomial Theorem, as above. However, the argument is exactly the same. That is,
$$
\lim_{n\to\infty}\left[\sum_{k=1}^nk^{-z}-\frac1{1-z}n^{1-z}-\frac12n^{-z}\right]
$$
converges uniformly to an analytic function for $\mathrm{Re}(z)\gt-1$. For $\mathrm{Re}(z)\gt1$, this function is easily seen to be $\zeta(z)$.
By analytic continuation, we get that this function is $\zeta(z)$ for $\mathrm{Re}(z)\gt-1$. For this question, plug in $z=-\frac12$.
