TITLE: Asymptotic behavior of the partial sums $\sum\limits_{k=1}^{n}k^{1/4} $
QUESTION [16 upvotes]: What is the asymptotic behavior of the sequence:
\begin{equation}
s_n=\sum_{k=1}^{n}k^{1/4}
\end{equation}
when $n\to \infty$?

REPLY [12 votes]: There is a standard technique that produces the complete asymptotic expansion for this sum and many others like it, which is to use harmonic sums and Mellin transforms.
Introduce the telescoping sum $$S(x) = \sum_{k\ge 1} \left(\sqrt[4]{k}-\sqrt[4]{x+k}\right).$$
This sum has the property that $$S(n) = \sum_{q=1}^n \sqrt[4]{q},$$ so that $S(n)$ is the value we are looking for.
Re-write the sum as follows:
$$S(x) = \sum_{k\ge 1} \sqrt[4]{k} \left(1-\sqrt[4]{x/k+1}\right).$$
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
$$\lambda_k = \sqrt[4]{k}, \quad \mu_k = \frac{1}{k} \quad \text{and} \quad
g(x) = 1 - \sqrt[4]{1+x}.$$
It follows that
$$\sum_{k\ge 1} \frac{\lambda_k}{\mu_k^s} = 
\sum_{k\ge 1} \sqrt[4]{k}\times k^s
= \zeta(-1/4-s)$$
which has fundamental strip $-1/4-s > 1$ or $s < -5/4.$
We need the Mellin transform $g^*(s)$ of $g(x)$ which is
$$\int_0^\infty \left(1 - \sqrt[4]{1+x}\right) x^{s-1} dx$$
which is immediately seen to be a beta function integral with value
$$g^*(s) = - \frac{1}{\Gamma(-1/4)} \Gamma(s)\Gamma(-1/4-s)$$
and fundamental strip $\langle -1, -1/4 \rangle,$ which is a problem, because the abscissa of convergence of the zeta function term lies outside this strip. Hence we need to shift $g^*(s)$ by using $$ g(x) = 1 + \frac{1}{4} x - \sqrt[4]{1+x}$$
with fundamental strip $\langle -2, -1 \rangle,$ which is good.
We have now added the following quantity to our sum:
$$\sum_{k\ge 1} \sqrt[4]{k} \frac{1}{4} \frac{x}{k}
= \frac{1}{4} x \zeta(3/4),$$
which we will have to remember to subtract from our final answer.
It follows that the Mellin transform $Q(s)$ of the harmonic sum $S(x)$ is given by
$$Q(s) =
- \frac{1}{\Gamma(-1/4)} \Gamma(s)\Gamma(-1/4-s)
\zeta(-1/4-s).$$
The Mellin inversion integral here is
$$\frac{1}{2\pi i} \int_{-3/2-i\infty}^{-3/2+i\infty} Q(s)/x^s ds$$
which we evaluate by shifting it to the right for an expansion at infinity.
First treat the pole from the zeta function term at $s=-5/4$, which has
$$\mathrm{Res}(Q(s)/x^s; s=-5/4) =
-\frac{1}{\Gamma(-1/4)} \Gamma(-5/4)\Gamma(1)\times -1 \times x^{5/4}
= -\frac{4}{5} x^{5/4}.$$
For the pole at $s=-1$ we get
$$\mathrm{Res}(Q(s)/x^s; s=-1) = -\frac{1}{4}\zeta(3/4)x.$$
We see that this is precisely the contribution that we added in when we shifted $g^*(s)$ into position and hence this residue will not be included in the asymptotic expansion.
For the pole at $s=-1/4$ from the compound gamma function term we obtain
$$\mathrm{Res}(Q(s)/x^s; s=-1/4) =
 -\frac{1}{\Gamma(-1/4)} \Gamma(-1/4) \times -1 \times\zeta(0)
= -\frac{1}{2} \sqrt[4]{x}.$$
For the pole at $s=0$ from the simple gamma function term we obtain
$$\mathrm{Res}(Q(s)/x^s; s=0) = -\frac{1}{\Gamma(-1/4)} \Gamma(-1/4) \zeta(-1/4)
= -\zeta(-1/4).$$
The remaining poles are at $s = q-1/4$ where $q\ge 1$ and contribute
$$\mathrm{Res}(Q(s)/x^s; s=q-1/4) =
-\frac{1}{\Gamma(-1/4)} \Gamma(q-1/4) \frac{(-1)^{q+1}}{q!} \zeta(-q) \frac{1}{x^{q-1/4}}
\\ = - \prod_{p=0}^{q-1} (p-1/4) \times
 \frac{(-1)^{q+1}}{q!} (-1)^q \frac{B_{q+1}}{q+1}  \frac{1}{x^{q-1/4}}
= \frac{1}{4^q} B_{q+1}
\frac{\prod_{p=0}^{q-1} (4p-1)}{(q+1)!}  \frac{1}{x^{q-1/4}}.$$
The zero values of the Bernoulli numbers correctly represent cancelation of the gamma function poles by the trivial zeros of the zeta function.
Setting $x=n$ and observing that the shift to the right produces a minus sign we obtain the following asymptotic expansion:
$$S(n) = \sum_{k=1}^n \sqrt[4]{k} \sim
\frac{4}{5} n^{5/4} + \frac{1}{2} \sqrt[4]{n} + \zeta(-1/4) 
- \sum_{q\ge 1} \frac{1}{4^q} B_{q+1}
\frac{\prod_{p=0}^{q-1} (4p-1)}{(q+1)!}  \frac{1}{n^{q-1/4}}.$$
Actually computing the Bernoulli number terms we get the expansion
$$\frac{4}{5} n^{5/4} + \frac{1}{2} \sqrt[4]{n} + \zeta(-1/4) +
1/48\,{n}^{-3/4}-{\frac {7}{15360}}\,{n}^{-11/4}+{\frac {11}{98304}}\,
{n}^{-{\frac {19}{4}}}\\-{\frac {4807}{62914560}}\,{n}^{-{\frac {27}{4}}
}+{\frac {13547}{134217728}}\,{n}^{-{\frac {35}{4}}}-{\frac {9360977}{
42949672960}}\,{n}^{-{\frac {43}{4}}}\\+{\frac {191649409}{274877906944}
}\,{n}^{-{\frac {51}{4}}}-{\frac {1089307862269}{351843720888320}}\,{n
}^{-{\frac {59}{4}}} +\cdots.$$
This MSE link points to a series of similar calculations.

Care must be taken with the asymptotics of these expansions since the Bernoulli numbers eventually outgrow all other terms. For example when $n=10$ the sum of the first $62$ Bernoulli terms converge to produce thirty digits of precision and diverge from then on. These expansions fit the definition however, where we say that an asymptotic sequence $\{\phi_k(n)\}$ has 
$$|\phi_{k+1}(n)/\phi_k(n)|\rightarrow 0$$ as $n\rightarrow\infty$ and
$$\lim_{n\rightarrow\infty}\frac{|S(n)-\sum_{k=1}^m\phi_k(n)|}{\phi_m(n)}=0.$$