TITLE: Asymptotic for sum
QUESTION [7 upvotes]: 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?

REPLY [4 votes]: Suppose  we  answer this  question  by  computing  the asymptotics  of
$$\sum_{k=1}^n \sqrt[3]{k}.$$ This is actually a textbook example of
harmonic summation techniques.
Introduce
$$S(x) = \sum_{k\ge 1} \left(\sqrt[3]{k}-\sqrt[3]{x+k}\right)$$
so that our answer is given by $S(n).$
Re-write $S(x)$ as follows:
$$S(x) = \sum_{k\ge 1} \sqrt[3]{k} \left(1-\sqrt[3]{x/k+1}\right).$$
The sum term is harmonic and  may be evaluated by inverting its Mellin
transform.
Recall the harmonic sum identity
$$\mathfrak{M}\left(\sum_{k\ge 1} \lambda_k g(\mu_k x);s\right) =
\left(\sum_{k\ge 1} \frac{\lambda_k}{\mu_k^s} \right) g^*(s)$$
where $g^*(s)$ is the Mellin transform of $g(x).$
In the present case we have
$$\lambda_k = \sqrt[3]{k}, \quad 
\mu_k = \frac{1}{k} \quad \text{and} \quad
g(x) = 1-\sqrt[3]{x+1}.$$
Canceling an  initial part  of the series  about zero of  the function
being transformed only  shifts the fundamental strip, so  we may start
by calculating the Mellin transform $h^*(s)$ of $h(x)=1-g(x)$ which is
$$\int_0^\infty \sqrt[3]{x+1}  x^{s-1} dx = 
\int_0^\infty \frac{x^{s-1}}{(1+x)^{s+(-1/3-s)}} dx
= \mathrm{B}(s, -1/3-s).$$

Now the convergence of  the beta function integral requires $\Re(s)>0$
and $\Re(-1/3-s)>0$  which is $-1/3>\Re(s)$ so this  does not converge
(half planes  do not intersect).   Therefore we shift  the fundamental
strip to $\langle -2, -1\rangle$ by canceling the two initial terms of
$\sqrt[3]{1+x}  \sim  1  +  \frac{1}{3}  x$  in  effect  using  $g(x)+
\frac{1}{3} x.$  (The reason why  we shifted to this  particular strip
will become clear later. We have now added
$$\sum_{k\ge 1} \sqrt[3]{k} \times \frac{1}{3} \frac{x}{k} = 
\frac{1}{3} x \sum_{k\ge 1} \frac{1}{k^{2/3}} = 
\frac{1}{3} \zeta(2/3) \times x$$
to $S(x)$ and will have to remember to subtract this quantity later.)

It follows that the Mellin transform of $g(x)$ is given by
$$g^*(s) = - \mathrm{B}(s, -1/3-s)
= - \frac{\Gamma(s)\Gamma(-1/3-s)}{\Gamma(-1/3)}
= -\frac{1}{\Gamma(-1/3)} \Gamma(s) \Gamma(-1/3-s).$$
Therefore the transform $Q(s)$ of $S(x)$ is
$$ Q(s) = -\frac{1}{\Gamma(-1/3)} \Gamma(s) \Gamma(-s-1/3) 
\zeta(-s-1/3)
\quad\text{because}\quad \sum_{k\ge 1} 
\frac{\lambda_k}{\mu_k^s} = 
\sum_{k\ge 1}\sqrt[3]{k} \times k^s 
= \zeta(-s-1/3).$$
The   half    plane   of   convergence    of   the   zeta    term   is
$\Re(s)<-4/3.$ Taking  into account  the intersection of  $\langle -2,
-1\rangle$  with the  half plane  of  convergence we  thus obtain  the
Mellin inversion integral
$$ S(x) = \frac{1}{2\pi i} 
\int_{-3/2-i\infty}^{-3/2+i\infty} Q(s)/x^s ds$$
which we evaluate  by shifting it to the right  for an expansion about
infinity.
We get $$\mathrm{Res}\left(Q(s)/x^s; s = -\frac{4}{3}\right) = 
-\frac{3}{4} x^{4/3}.$$
(Not to worry  about the apparent sign error,  since the residues will
be negated because we are shifting to the right.)

We also have
$$\mathrm{Res}\left(Q(s)/x^s; s = -1\right) = 
-\frac{1}{3} \zeta(2/3) \times x,$$
and this will not be contributing as explained earlier.

Furthermore we have
$$\mathrm{Res}\left(Q(s)/x^s; s = -\frac{1}{3}\right) = 
-\frac{1}{2} x^{1/3}.$$

Continuing we obtain
$$\mathrm{Res}\left(Q(s)/x^s; s = 0\right) = 
-\zeta(-1/3).$$

The last pole to merit special treatment is the one at $s=2/3$
where we obtain
$$\mathrm{Res}\left(Q(s)/x^s; s = \frac{2}{3}\right) = 
-\frac{1}{36} x^{-2/3}.$$

For the remaining poles at $s=q-1/3$ where $q\ge 2$ we obtain
$$\mathrm{Res}\left(Q(s)/x^s; s = q-1/3\right) = 
\frac{\Gamma(q-1/3)}{\Gamma(-1/3)} \times \frac{(-1)^q}{q!}
\times \zeta(-q) \times x^{1/3-q}.$$
This is
$$(-1)^q
\frac{\prod_{k=0}^{q-2} (3k+2)}{3^q \times q!} 
\frac{B_{q+1}}{q+1} \times x^{1/3-q}$$
or
$$(-1)^q \times B_{q+1} \times
\frac{\prod_{k=0}^{q-2} (3k+2)}{3^q \times (q+1)!}
\times x^{1/3-q}.$$

Now this only contributes when $q+1$ is even 
so that we may simplify the entire contribution to 
$$- \sum_{p\ge 1} B_{2p+2} 
\frac{\prod_{k=0}^{2p-1} (3k+2)}{3^{2p+1} \times (2p+2)!}
\times x^{-2/3-2p}.$$

This gives for the asymptotic expansion
$$S(x) \sim \frac{3}{4} x^{4/3}
+ \frac{1}{2} x^{1/3}
+ \zeta(-1/3)
\\+ \frac{1}{36} x^{-2/3}
+ \sum_{p\ge 1} B_{2p+2} 
\frac{\prod_{k=0}^{2p-1} (3k+2)}{3^{2p+1} \times (2p+2)!}
\times x^{-2/3-2p}.$$

We have in particular with $x=n$ the asymptotic expansion
$$3/4\,{n}^{4/3}+1/2\,\sqrt [3]{n}+\zeta  \left( -1/3 \right) +1/36
\,{n}^{-2/3}\\-{\frac {1}{1944\,{n}^{8/3}}}+{\frac {11}{91854\,{n}^
{14/3}}}-{\frac {187}{2361960}{n}^{-{\frac {20}{3}}}}+{\frac {
1955}{19131876}{n}^{-{\frac {26}{3}}}}-\cdots$$

There is a very similar calculation at this
MSE link.