Sum of 1.5-powers of natural numbers

I recently have met the following approximate equation:

$$\sum_{k=1}^n k^{1.5}\approx\frac{n^{2.5}+(n+1)^{2.5}}{5}.$$

It's a rather accurate approximation (for $n=40$ the absolute error is $\approx 1.67$ and it increases very slowly), and looks elegant, so I immediately tried to prove it by expanding the right-hand side using binomial expansion. What I got was

$$\frac25n^{2.5}+\frac12n^{1.5}+\frac{15}8n^{0.5}+\dots,$$

nothing particularly resemblant to the left-hand side. After that I went to googling for something about sums of powers of natural numbers, found Faulhaber's formula

$$\sum_{k=1}^n k^p = \frac{1}{p+1}\sum_{j=0}^p (-1)^j \genfrac{(}{)}{0}{}{p+1}{j}B_j n^{p+1-j},$$

and after (mis)using it with $p=1.5$ I obtained $$\frac25n^{2.5}+\frac12n^{1.5}.$$

While this last result is a more accurate approximation for $\sum_{k=1}^n k^{1.5}$, and is a partial sum for the expansion of the original right-hand side, I still can't understand how the original approximation was obtained. Can someone suggest its source, how one could have conceived of it?

Solution 1:

We have the inequalities

$$\int_0^nx^{3/2}dx<\sum_{k=1}^n k^{3/2}<\int_0^{n+1}x^{3/2}dx$$

whereupon carrying out the integrals yields

$$\frac25 n^{5/2}<\sum_{k=1}^n k^{3/2}<\frac25 (n+1)^{5/2}$$

The approximation of interest is simply the average of the upper and lower limits of the sum and is expressed as

$$\bbox[5px,border:2px solid #C0A000]{\sum_{k=1}^n k^{3/2}\approx \frac{n^{5/2}+(n+1)^{5/2}}{5}}$$

A much better approximation is found using the Euler-Maclaurin Formula and is

$$\bbox[5px,border:2px solid #C0A000]{\sum_{k=1}^n k^{3/2}=\frac25n^{5/2}+\frac12 n^{3/2}+\frac18n^{1/2}+C+\frac{1}{1920}n^{-3/2}+O(n^{-7/2})}$$

where the constant $C$ can be found numerically and is approximately given by $C\approx -0.025496493$.

Solution 2:

$${n^{2.5}+(n+1)^{2.5}\over5}={n^{2.5}\over5}\left(1+\left(1+{1\over n}\right)^{2.5} \right)$$

and

$$\left(1+{1\over n}\right)^{2.5}\approx1+{2.5\over n}$$

so

$${n^{2.5}+(n+1)^{2.5}\over5}\approx{n^{2.5}\over5}\left(2+{2.5\over n}\right)={2\over5}n^{2.5}+{1\over2}n^{1.5}$$