# How to calculate the asymptotic expansion of $\sum \sqrt{k}$?

Denote $u_n:=\sum_{k=1}^n \sqrt{k}$. We can easily see that $$k^{1/2} = \frac{2}{3} (k^{3/2} - (k-1)^{3/2}) + O(k^{-1/2}),$$ hence $\sum_1^n \sqrt{k} = \frac{2}{3}n^{3/2} + O(n^{1/2})$, because $\sum_1^n O(k^{-1/2}) =O(n^{1/2})$.

With some more calculations, we get $$k^{1/2} = \frac{2}{3} (k^{3/2} - (k-1)^{3/2}) + \frac{1}{2} (k^{1/2}-(k-1)^{-3/2}) + O(k^{-1/2}),$$ hence $\sum_1^n \sqrt{k} = \frac{2}{3}n^{3/2} + \frac{1}{2} n^{1/2} + C + O(n^{1/2})$ for some constant $C$, because $\sum_n^\infty O(k^{-3/2}) = O(n^{-1/2})$.

Now let's go further. I have made the following calculation $$k^{1/2} = \frac{3}{2} \Delta_{3/2}(k) + \frac{1}{2} \Delta_{1/2}(k) + \frac{1}{24} \Delta_{-1/2}(k) + O(k^{-5/2}),$$ where $\Delta_\alpha(k) = k^\alpha-(k-1)^{\alpha}$. Hence : $$\sum_{k=1}^n \sqrt{k} = \frac{2}{3} n^{3/2} + \frac{1}{2} n^{1/2} + C + \frac{1}{24} n^{-1/2} + O(n^{-3/2}).$$ And one can continue ad vitam aeternam, but the only term I don't know how to compute is the constant term.

How do we find $C$ ?

-
you want to calculate the taylor series of a sequence? for taylor you need a differentiable function ... –  Dominic Michaelis Mar 10 '13 at 17:44
When you write: $\sum_n^\infty O(k^{-3/2}) = O(k^{-1/2})$ you presumably mean $O(n^{-1/2})$? –  Thomas Andrews Mar 10 '13 at 17:44
The $\sum_{1 \le k \le n} O(k^{-1/2}) = O(n^{-1/2})$ gives me the willies... to start with, the largest term of the sum $\sum_{1 \le k \le n} k^{-1/2}$ is $O(1)$. If anything, express the missing terms more precisely, and estimate (or bound) $\sum_{1 \le k \le n} f(k)$ with $\int_1^n f(x) dx$. Or use Euler-Maclaurin's formula directly –  vonbrand Mar 10 '13 at 18:32
@vonbrand. I have written $O(n^{+1/2})$ not $O(n^{-1/2})$. Euler-Maclaurin tells : $\int_1^n \sqrt{t} dt = \frac{1}{2}\sqrt{1} + \dots + \sqrt{1/2} \sqrt{n} + \frac{B_2}{2}(\frac{1}{\sqrt{n}}-\frac{1}{2}) + \int_1^n \frac{P_2(t)}{2} \frac{-1/4}{t^{3/2}} dt$. But $\int_1^n \frac{P_2(t)}{2} \frac{-1/4}{t^{3/2}} dt = Constant + O(1/n^{1/2})$, so this method doesn't give $C$ (or am I misusing it ?) –  Auguste Hoang Duc Mar 10 '13 at 20:15
$C\ne 0$. $C=\zeta(-\frac 12)\approx-0.207886224977$. –  David Moews Mar 11 '13 at 3:02