10
$\begingroup$

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?

$\endgroup$
1
  • 1
    $\begingroup$ Off the top of my head, I would guess that it corresponds to treating the sum as an approximation to the integral $\int_0^{n+1}n^{3/2}$ and using the appropriate approximate error term. Once one has the $\frac25n^{5/2}+\frac12n^{3/2}$ approximation from whatever source, too, it's possible to recognize the latter as an approximate finite difference between $(n+1)^{5/2}$ and $n^{5/2}$ and find the right scale. $\endgroup$ Aug 12, 2015 at 0:50

2 Answers 2

12
$\begingroup$

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 $C\approx -0.025496493$.

$\endgroup$
4
  • 1
    $\begingroup$ Nice! This is the integral approximation I was looking for but couldn't find. (Although I think you want zero rather than one as a lower bound for the integrals; the math comes out much easier.) $\endgroup$ Aug 12, 2015 at 6:14
  • 1
    $\begingroup$ for OP: note that the two estimates come from upper-bounding the left integral by chopping into the intervals $[0,1], [1, 2], \ldots, [n-1, n]$ and using the fact that $x^{3/2}\leq k^{3/2}$ for $x\in[k-1,k]$, and similarly lower-bounding the right integral by chopping into the intervals $[0,1], [1,2], \ldots, [n, n+1]$ and using the fact that $x^{3/2}\geq k^{3/2}$ for $x\in [k, k+1]$. $\endgroup$ Aug 12, 2015 at 6:15
  • $\begingroup$ A-ha! It's "replace sum with trapezoid integration" trick, and then average the under- and over-estimated values. Thanks, now I see it. And yeah, that explains why it's less accurate than $\frac25n^{5/2}+\frac12n{3/2}$—it's still a first-order approximation. $\endgroup$
    – Joker_vD
    Aug 12, 2015 at 19:08
  • 1
    $\begingroup$ @Joker_vD You're welcome! My pleasure. And it's even simpler -no trapezoidal trick at all. Instead, we see that the sum is (i) the right-hand side Riemann Sum for $\int_0^n x^{3/2}dx$ and (ii) the left-hand side Riemann Sum for for $\int_0^{n+1} x^{3/2}dx$ . And since $x^{3/2}$ monotonically increases, the sum is an upper bound for the former and a lower bound for the latter integral. Then, the approximation simply averages the upper and lower bounds. That said, I encourage you to check out the Euler-Maclaurin Sum Formula as it is a powerful tool to be sure. $\endgroup$
    – Mark Viola
    Aug 12, 2015 at 20:06
1
$\begingroup$

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

and

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

so

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

$\endgroup$
2
  • 2
    $\begingroup$ But why is this equal to the sum of $k^{1.5}$? $\endgroup$
    – Joker_vD
    Aug 12, 2015 at 4:52
  • $\begingroup$ @Joker_vD, perhaps I misunderstood the nature of your question. I thought you were asking how to get from one approximation to the other. In any event, Dr. MV has given a better, more complete answer. $\endgroup$ Aug 12, 2015 at 12:37

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .