2
$\begingroup$

Possible Duplicate:
Finite Sum of Power?

Is there a general expression for $\sum_{k=1}^n k^x$ for any integer value of $x$? The table for $x=1,2,\dots 10$ is given here. Is there formula for any value of $x$?

$\endgroup$

marked as duplicate by Martin Sleziak, Sasha, sdcvvc, William, J. M. is a poor mathematician Sep 3 '12 at 23:49

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ This seems to be the same question as this one. $\endgroup$ – Martin Sleziak Aug 31 '12 at 10:28
3
$\begingroup$

For arbitrary natural $x\in\mathbb{N}$, the general formula is given by Faulhaber's Formula, which is $$\sum_{k=1}^n k^x = \frac{1}{x+1} \sum_{i=0}^x (-1)^i{x+1 \choose i} B_i \cdot n^{x+1-i}$$ where $B_i$ are the Bernoulli Numbers with $B_1 = -\frac{1}{2}$.

I am not too convinced that there exists a nice closed form expression for arbitrary $x$.

$\endgroup$
  • $\begingroup$ that's more scarier :(( ... is there any formula without summation? $\endgroup$ – hasExams Aug 31 '12 at 3:30
  • $\begingroup$ @testuser If there is one, then I don't know about it. But I really doubt that there exists anything much nicer than this expression. You can already see in the list you provided that the expressions are getting quite large by the $10$ sum. This is simply a more compact way of expressing the inevitably larger sums that occur with higher powers. $\endgroup$ – EuYu Aug 31 '12 at 3:32
  • 3
    $\begingroup$ Technically there is a "formula" without summation using the Bernoulli polynomials, specifically $$\sum_{k=0}^{x} k^p = \frac{B_{p+1}(x+1)-B_{p+1}(0)}{p+1},$$ but in some sense that is cheating, because the Bernoulli polynomial by definition is just the above sum! $\endgroup$ – Eric Naslund Aug 31 '12 at 3:40
  • 1
    $\begingroup$ @testuser Of course the right-hand-side is related to Bernoulli polynomials $$ \sum_{k=1}^n k^m = \frac{1}{m+1} \left(B_{m+1}(n+1)-B_{m+1}(1)\right)$$ $\endgroup$ – Sasha Aug 31 '12 at 3:40
  • 1
    $\begingroup$ Also the sum defines what is known as generalized harmonic numbers: $$ \sum_{k=1}^n k^m = H_n^{(-m)} $$ This is somewhat a tautology, though. $\endgroup$ – Sasha Aug 31 '12 at 3:42
2
$\begingroup$

I already answered a question similar to this on this website. See here. You can use the zeta function and the Hurwitz zeta function to get a closed form formula to your sum,

$$ \sum_{k=1}^n k^x = \zeta(-x) + \zeta(-x,n+1) \,. $$

See reference 1 and reference 2.

$\endgroup$
  • $\begingroup$ Funny how that analytic continuation stuff works here. ;) $\endgroup$ – Simply Beautiful Art Nov 29 '16 at 1:10

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