Asymptotic for sum

Question

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?

Answer 1

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.

Answer 2

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

$$\sum_{k=1}^{n}k^{1/3} = \frac{3}{4}\left(n^{4/3}-1\right) + \frac1{2}[n^{1/3}-1]+\frac{1}{36}[n^{-2/3}-1] - \frac{1}{1944}[n^{-8/3}-1] \\ +\frac{11}{91854}[n^{-14/3}-1]+ \ldots$$

Answer 3

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