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For example, the sum of n is n(n+1)/2, the dominating term is n square(let say this is order 2). For the sum of n^2, the order is 3. Then for the sum of n^k, is the order k+1?

I been searching Faulhaber's formula and Bernoulli numbers, I'm not sure what is the order of it. It's much more complicated than i think.

Any ideas? Thanks in advance

Checkman
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2 Answers2

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By the definition of Riemann integral, $\frac1n \sum_{k=1}^n (\frac{k}{n})^\alpha \rightarrow \int_0^1 t^\alpha dt=\frac1{\alpha+1}$.

Then $\sum_{k=1}^n k^\alpha \sim \frac{n^{\alpha+1}}{\alpha+1}$

D.L.
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You are correct. Let the function $F_k$ be defined as $$ F_k(n) = \sum_{i = 1}^n i^k $$ and assume that $F_k(n)$ is a polynomial of some degree, say $$ F_k(n) = a_0 + a_1n + \cdots + a_ln^l $$ Then the function $f_k(n) = F(n) - F(n-1) = n^k$ is a polyniomial of order $k$. But that means that we must have $l = k+1$ (and $a_l = a_{k+1} = 1/(k+1)$).

Of course, what's left is to show that $F_k(n)$ is indeed a polynomial for all $k$.

Arthur
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  • I remember that with this method you can even have an exact formula by recurrence on $k$. Isn't it ? (I guess I have forgotten how :( ). – D.L. Sep 14 '15 at 07:38
  • Nope I was confused, it was something else... I will post about it soon^^ – D.L. Sep 14 '15 at 07:40
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    With your notations, by using $(k+1)^p-k^p=\sum_{j=0}^{p-1} \binom{p}{j}k^j$ and summing for k from 0 to n, you get :

    $(n+1)^p = \sum_{j=0}^{p-1} \binom{p}{j} F_j(n)$, and then you can get the $F_k$ by recurrence.

     – D.L. Sep 14 '15 at 07:43
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    @D.L. And this would prove that $F_k$ is a polynomial pretty easily. – Arthur Sep 14 '15 at 07:52