6
$\begingroup$

When I was taking a shower this problem came into my mind...

Let $f(n, s) = 1^{-s} + 2^{-s} + 3^{-s} + \cdots + n^{-s}$ be the partial sum of the $\zeta$ function.

In the cases where $s$ is a negative integer, there is the usual closed-form formula for $f(n,s)$ involving Bernoulli numbers.

However what about the cases in which $s$ is a positive integer?

For example, when $s = 1$, we have

$f(1,1) = 1$

$f(2,1) = \frac 1 1 + \frac 1 2 = \frac 3 2$

$f(3,1) = \frac 1 1 + \frac 1 2 + \frac 1 3 = \frac {11} 6$

$\cdots$

$f(10, 1) = \frac 1 1 + \frac 1 2 + \frac 1 3 + \cdots + \frac 1 {10} = \frac{7381}{2520}$

Can one say anything about prime factors of the numerator and the denominator, in the final simplified fraction?

And what if $s$ is some larger integer?

Thank you very much.

p.s. A non-related beautiful paper on partial sums of $\zeta$, found when I tried to google the answer for my question: www.cecm.sfu.ca/~pborwein/MITACS/papers/borwein.ps

$\endgroup$
3
  • 6
    $\begingroup$ The partial sums of the harmonic series are named harmonic numbers. Check en.wikipedia.org/wiki/Harmonic_number, and the references therein. Also, of course, "Concrete Mathematics" by Graham Knuth Patashnik has a lot material about the subject. $\endgroup$ Jun 26, 2010 at 8:00
  • $\begingroup$ A plenty of stuff can be found at usna.edu/Users/math/meh/biblio.html under the rubric "G. FINITE MULTIPLE HARMONIC SUMS". $\endgroup$ Jun 26, 2010 at 8:32
  • $\begingroup$ In the appendix of Michael Artin's Algebra, a closed form formula for the partial sum of the harmonic series is asked, and I still wonder what simple function he had in mind...the digamma function is not simple, I think. $\endgroup$
    – Unknown
    Oct 30, 2010 at 12:38

4 Answers 4

6
$\begingroup$

These are called the harmonic numbers. There is a lot of information about them at http://mathworld.wolfram.com/HarmonicNumber.html The numerators are discussed at http://oeis.org/A001008 and the denominators at http://oeis.org/A002805

$\endgroup$
0
1
$\begingroup$

Denominator is quite large (for example, it is divisible by all prime powers from $[n/2,n]$, by greatest power of 2, not exceeding $n$, for given $c>0$ it is divisible by all primes from $[cn,n]$ provided $n$ is lage enough; and so on). As for numerator, I used to think about it and get almost nothing. The only very weak, but however non-trivial statement about it I was able to prove is that for infinitely many values of $n$ the numerator is not a power of prime (http://www.artofproblemsolving.com/Forum/viewtopic.php?f=59&t=1778).

$\endgroup$
1
$\begingroup$

There is a closed form for what you are asking for. The partial sum is

$$\sum_x x^{-s} =\frac{(-1)^{s-1}\psi^{(s-1)}(x)}{\Gamma(s)}+C$$

for any natural s.

$\endgroup$
4
  • $\begingroup$ Did you discover this yourself, or is there a reference? $\endgroup$
    – S. Carnahan
    Dec 20, 2010 at 17:13
  • $\begingroup$ It is a known formula. One can find it in many sources. For example, here: mathworld.wolfram.com/PolygammaFunction.html (formula 12) $\endgroup$
    – Anixx
    Dec 20, 2010 at 19:49
  • 3
    $\begingroup$ @Scott: polygamma functions and generalized harmonic numbers are essentially the same banana. $\endgroup$ Dec 20, 2010 at 22:01
  • $\begingroup$ Even more: zeta, polygamma, harmonic numbers, Bernoulli polynomials and even logarithmic integral are essentially the same banana... $\endgroup$
    – Anixx
    Dec 21, 2010 at 5:21
0
$\begingroup$

It is sometimes proposed that if $n$ is the number of words in a language then the relative frequency of the $k$th-most frequently used word is $$ \frac{k^{-s}}{\sum_{j=1}^n j^{-s}}, $$ where $s$ is a parameter that depends on which language it is. George Kingsley Zipf famously proposed $s=1$ and some sources say $s$ should be just slightly bigger than 1. I've never heard that any of this is really well supported by empirical evidence, but maybe I missed that. If you look around the internet you'll see some people getting quite enthusiastic about this.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.