How to estimate $ \left(1 + \sqrt{2} + \sqrt{3} + \dots + \sqrt{n} \right) - \frac{2}{3} n \sqrt{n}$?

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 ($<5\cdot 10^{-4}$).

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 $z=-\frac12$.