Is there a way to calculate a sum of non-integer positive powers of integers?

$\sum_{k=1}^nk^p: n \in \mathbb{N}, p \in \mathbb{R^+}$

There's a Faulhaber's formula, but as far as I can see, it is applicable only to integer powers. Is it possible to generalize it w/o getting too complex computation?

The thing is the formula needs to be calculated on a computer and if a solution involves calculating integrals up to infinity it might be simpler to calculate the sum directly.

$n$ might be up to $10^{12}$, approximations are also an option.


This is related to the (Hurwitz) Zeta function or generalized harmonic numbers $$\sum_{k=1}^nk^p = \zeta(-p) -\zeta(-p, 1 + n) = H_n^{(-p)}$$

Example: for $n=10^9, p=0.1$ I get with double precision functions the value $7.2211657737752\cdot 10^9$ or for $n=10^{12},\; p=1/4$ you have $2.8109134757068\cdot 10^{13}$

These numbers are computed with my Pascal routines, you can find C functions in the GSL or Cephes libraries, with Python there is mpmath.

  • $\begingroup$ I feel like this just comes from the definition of those special functions though. $\endgroup$ – Ethan Nov 2 '17 at 16:34
  • $\begingroup$ @Ethan: Yes of course, but you can use them to actually compute the sums without adding $10^9$ terms, see my example. $\endgroup$ – gammatester Nov 2 '17 at 16:38

For integer $p$, the first Faulhaber terms are


For large $n$, these terms are quickly decreasing and I wouldn't be surprised that you can simply plug fractional values of $p$ to get precise estimates.

(I have no serious justification.)

  • 1
    $\begingroup$ If you add the three terms you get an error of $0.417$ for my $10^p, 0.1$ example, i.e. relative error of $0.58\cdot10^{-10}.$ A justificatiion might come from the representations by the Euler–Maclaurin formula dlmf.nist.gov/25.2#iii and dlmf.nist.gov/25.11#iii $\endgroup$ – gammatester Nov 2 '17 at 19:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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