Asymptotic behavior of the partial sums $\sum\limits_{k=1}^{n}k^{1/4} $ What is the asymptotic behavior of the sequence:
\begin{equation}
s_n=\sum_{k=1}^{n}k^{1/4}
\end{equation}
when $n\to \infty$?
 A: Euler McLaurin Summation yields:
$$\sum_{k=0}^{n} k^{1/4} = \frac{4}{5} n^{5/4} + \frac{1}{2} n^{1/4} + C + \mathcal{O}(n^{-3/4})$$
It can be shown that $C = \zeta(-1/4)$ where $\zeta$ is the Riemann-Zeta function as defined on the whole plane.
Note that this is a much stronger statement than $\sum_{k=0}^{n}k^{1/4} \sim \frac{4}{5} n^{5/4}$, which (in the current context) means that
$$\lim_{n \to \infty} \frac{\sum_{k=0}^{n}k^{1/4}}{\frac{4}{5} n^{5/4}} = 1$$
What Euler Mclaurin gives us is the following:
$$\lim_{n \to \infty} \sum_{k=0}^{n}k^{1/4}-  \frac{4}{5} n^{5/4} - \frac{1}{2} n^{1/4} - C = 0 $$
Which actually implies that
$$\lim_{n \to \infty} \sum_{k=0}^{n}k^{1/4}-  \frac{4}{5} n^{5/4} = \infty$$
There is also an elementary proof of this fact (but which does not determine the exact value of $C$) here: How closely can we estimate $\sum_{i=0}^n \sqrt{i}$
A: Let $k\geqslant1$. Since the function $x\mapsto x^{1/4}$ is increasing, $(k-1)^{1/4}\leqslant x^{1/4}\leqslant k^{1/4}$ for every $k-1\leqslant x\leqslant k$. Integrating this double inequality yields
$$
(k-1)^{1/4}\leqslant\int_{k-1}^kx^{1/4}\mathrm dx\leqslant k^{1/4}.
$$
Summing these from $k=1$ to $k=n$ yields
$$
s_{n-1}=\sum_{k=0}^{n-1}k^{1/4}\leqslant\int_{0}^nx^{1/4}\mathrm dx\leqslant \sum_{k=1}^{n}k^{1/4}=s_n.
$$
Since the integral is $\tfrac45n^{5/4}$ and $s_{n-1}=s_n-n^{1/4}$, this yields, for every $n\geqslant1$,
$$
\tfrac45n^{5/4}\leqslant s_n\leqslant\tfrac45n^{5/4}+n^{1/4}.
$$
Note finally that $n^{1/4}=o(n^{5/4})$ hence a (much weakened) version of this is $s_n\sim\tfrac45n^{5/4}$.
A: Here is a diagram to accompany Did's fine answer:

