Asymptotic for sum How can I find formula for
$\displaystyle{\sqrt[3]1 + \sqrt[3]2 + \sqrt[3]3 + \cdots + \sqrt[3]n}$
with an accuracy ${\rm O}\left(\, 1 \over \vphantom{\LARGE A}n^{5}\,\right)$
Is here we should use Bernoulli polynomials?
 A: Use Euler-Maclaurin summation with $f(x) = x^{1/3}$:
$$\sum_{k=1}^{n}k^{1/3} = \int_{1}^{n}f(x)\,dx + \frac1{2}[f(n)-f(1)]+\frac{B_2}{2!}[f'(n)-f'(1)] + \frac{B_4}{4!}[f'''(n)-f'''(1)]+ \ldots +\frac{(-1)^mB_m}{m!}[f^{(m-1)}(n)-f^{(m-1)}(1)] + R_m(n)$$
where the remainder is
$$R_m(n) = \frac{(-1)^{m+1}B_m}{m!}\int_1^nB_m(\{x\})f^{(m)}(x) \, dx$$
A bound on the error term $R_m(n)$ is
$$R_m(n) = \theta\frac{B_{m+2}}{(m+2)!}[f^{(m+1)}(n)-f^{(m+1)}(1)].$$
Note that 
$$f^{(m)}(n) = O[n^{-(3m-1)/3}].$$
We have $B_n = 0$ for odd $n\geq 3$, and $B_2 = 1/6$, $B_4 = -1/30$, $B_6 = 1/42, ...$
Then
$$\sum_{k=1}^{n}k^{1/3} = \frac{3}{4}\left(n^{4/3}-1\right) + \frac1{2}[n^{1/3}-1]+\frac{1}{36}[n^{-2/3}-1] - \frac{1}{1944}[n^{-8/3}-1] \\ +\frac{11}{91854}[n^{-14/3}-1]+ \ldots$$
A: You may start with the exact value of $\,\zeta\left(-\dfrac 13\right)\,$ and remove the correction terms from the Euler-Maclaurin expansion of $\zeta$ as provided at the end of this answer (i.e. set $z=-\dfrac 13$ there !).
or use directly this pari/gp script to compute $f(-1/3,n)$ :
f(x,n)=zeta(x)-(+1/((x-1)*n^(x-1))-1/(2*n^x)+x/(12*n^(x+1))-x*(x+1)*(x+2)/(720*n^(x+3))+x*(x+1)*(x+2)*(x+3)*(x+4)/(30240*n^(x+5)))  
g(n)=sum(k=1,n,k^(1/3))
d(n)=f(-1/3,n)-g(n)
> \p 48
> d(1000)  
%9 = 7.9171433319144741701079982817 E-25

