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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

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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. –  Pietro Majer Jun 26 '10 at 8:00
    
A plenty of stuff can be found at usna.edu/Users/math/meh/biblio.html under the rubric "G. FINITE MULTIPLE HARMONIC SUMS". –  Wadim Zudilin Jun 26 '10 at 8:32
    
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. –  Unknown Oct 30 '10 at 12:38
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4 Answers

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

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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).

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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.

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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.

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Did you discover this yourself, or is there a reference? –  S. Carnahan Dec 20 '10 at 17:13
    
It is a known formula. One can find it in many sources. For example, here: mathworld.wolfram.com/PolygammaFunction.html (formula 12) –  Anixx Dec 20 '10 at 19:49
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@Scott: polygamma functions and generalized harmonic numbers are essentially the same banana. –  J. M. Dec 20 '10 at 22:01
    
Even more: zeta, polygamma, harmonic numbers, Bernoulli polynomials and even logarithmic integral are essentially the same banana... –  Anixx Dec 21 '10 at 5:21
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