Euler-Maclaurin Summation Using EM summation formula estimate 
$$
\sum_{k=1}^n \sqrt k
$$
up to the term involving $\frac{1}{\sqrt n}$
My attempt is 
$$
\sum_{k=1}^n \sqrt k  = \frac{2 \sqrt{n^3}}{3} -\frac{2}{3}  + \frac 1 2 (\sqrt n -1)+ \frac{1}{24} (\frac{1}{\sqrt n} -1) + \int_1^n P_{2k+1}(x)f^{(2k+1)}(x)dx
$$
I am not sure what can be said about the integral. Please tell me if I have made a mistake and how I can solve the integral? Have I stopped the summation at the right point?
 A: Using the Euler-Maclaurin Sum Formula, we get
$$
\sum_{k=1}^n\sqrt{k}=\tfrac23n^{3/2}+\tfrac12n^{1/2}+\zeta(-\tfrac12)+\tfrac1{24}n^{-1/2}-\tfrac1{1920}n^{-5/2}+\tfrac1{9216}n^{-9/2}+O(n^{-13/2})
$$
where the constant $\zeta(-\frac12)=-0.20788622497735456602$ is explained below.

By the Euler-Maclaurin Sum Formula, we have for $\mathrm{Re}(s)\gt-1$
$$
\sum_{k=1}^nk^{-s}=\frac1{1-s}n^{1-s}+\frac12n^{-s}+\zeta_\ast(s)+O(n^{-s-1})\tag{1}
$$
Define the sequence of functions
$$
\zeta_n(s)=\sum_{k=1}^nk^{-s}-\frac1{1-s}n^{1-s}-\frac12n^{-s}\tag{2}
$$
For all $n\ge1$, $\zeta_n(s)$ is analytic. For $\mathrm{Re}(s)\gt1$, $\lim\limits_{n\to\infty}\zeta_n(s)=\zeta(s)$. Estimate $(1)$ says that on compact subsets of $\{s\in\mathbb{C}\setminus\{1\}:\mathrm{Re(s)\gt-1}\}$, $\zeta_n(s)$ converges uniformly. Thus, for $s\in\mathbb{C}\setminus\{1\}$ and $\mathrm{Re}(s)\gt-1$, $\lim\limits_{n\to\infty}\zeta_n(s)$ is analytic, so by analytic continuation, for $\mathrm{Re}(s)\gt-1$,
$$
\zeta_\ast(s)=\lim\limits_{n\to\infty}\zeta_n(s)=\zeta(s)\tag{3}
$$
