Possible Duplicate:
why is $\sum\limits_{k=1}^{n} k^m$ a polynomial with degree $m+1$ in $n$

Any hints that can take me from here or am I completely lost.





Where S(n,a) is a stirling number of second kind.


marked as duplicate by Aryabhata, Did, anon, Eric Naslund, t.b. Apr 12 '12 at 11:21

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.

  • 4
    $\begingroup$ I assume $p$ is a positive integer. There is a moderately complicated General Formula for the sum, involving the Bernoulli numbers. The term "power series" is a technical term that refers to something else. $\endgroup$ – André Nicolas Apr 12 '12 at 5:39
  • $\begingroup$ Yes p is a positive integer. Thanks for correcting me. $\endgroup$ – bspk Apr 12 '12 at 5:44
  • $\begingroup$ If you like you look at go.helms-net.de/math/potenzsummen/potenzsummen_1.htm which is a very old and completely elementary treatize of mine, of when I explored this myself the first time. Unfortunately in german, but I think the formulae and expressions are clear enough to hint you to a fruitful direction (for instance introducing how the Eulerian numbers of hkju's answer come into play) $\endgroup$ – Gottfried Helms Apr 12 '12 at 9:50

$$\sum_{k=1}^n k^p = \sum_{k=0}^{p-1} T(p,k)\binom{n+1+k}{p+1}. \mbox{ (hence, its degree is $p+1$)}$$, where $T(p,k)$ is the Eulerian number (cf.OEIS A008292). For example, $p=2$:$$\sum_{k=1}^n k^2 = \sum_{k=0}^{1} T(2,k)\binom{n+1+k}{3}=(1)\binom{n+1}{3}+(1)\binom{n+2}{3}=\frac{n(n+1)(2n+1)}{6}$$ $p=3$:$$\sum_{k=1}^n k^3 = \sum_{k=0}^{2} T(3,k)\binom{n+1+k}{4}=(1)\binom{n+1}{4}+(4)\binom{n+2}{4}+(1)\binom{n+3}{4}=\frac{n^2 (n+1)^2}{4}$$


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