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Denote $F_k(n)$ the sum of all $k$-th powers of first $n$ natural numbers. Also known as Faulhauber's formula. Let's give few examples: $$F_1(n)=1+2+3+...+n=\frac{n(n+1)}{2}$$ $$F_2(n)=1+4+9+...+n^2=\frac{n(n+1)(2n+1)}{6}$$ Now we know that the Riemann Zeta function is defined as follows for $s>1$ $$\zeta(s)=\sum_{n=1}^{\infty}\frac{1}{n^s}$$ From the analytic continuation we know that $\zeta(-2n)=0$ for all natural $n$. Also $\zeta(-n)=(-1)^n \frac{B_{n+1}}{n+1}$ where $B_n$ is $n$-th Bernoulli number. Now what i found was, that if take the definite integral from $-1$ to $0$ I get the riemann zeta function at $-k$: $$\int_{-1}^{0}{F_k(n)}dn=\zeta(-k)$$ Let me give few examples: $$\int_{-1}^{0}{F_1(n)}dn=\int_{-1}^{0}{\frac{n^2}{2}+\frac{n}{2}}dn=-\frac{1}{12}$$ $$\int_{-1}^{0}{F_2(n)}dn=\int_{-1}^{0}{\frac{n^3}{3}+\frac{n^2}{2}+\frac{n}{6}}dn=0$$ and so on... Now i know these sums of powers are somehow related to the Bernoulli numbers and so are the values of the riemann zeta function at negative integers. But i can't seem to find connection between these. Would anyone please give me an argument, why this holds?

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You can do it using the Faulhaber's polynomials, the binomial series, and some analytic continuation.

By induction we have the polynomials

$$F_k(N) = \sum_{n=1}^N n^k = \sum_{m=0}^{k+1} c_{m,k} N^m$$

For $|x| < 1$ we have the Taylor series $$(1+x)^{-s} = \sum_{l=0}^\infty {-s \choose l} x^l, \qquad {-s \choose l} = \prod_{j=0}^{l-1} \frac{-s-j}{j+1}, \\ n^{-s} - (n+1)^{-s} = n^{-s} (1-(1+n^{-1})^{-s})=-n^{-s} \sum_{l=1}^\infty {-s \choose l} n^{-l}$$ At first for $\Re(s) > k+1$ and by analytic continuation for every $s$ $$\zeta(s-k) = \sum_{n=1}^\infty n^k n^{-s} = \sum_{n=1}^\infty F_k(n) (n^{-s}-(n+1)^{-s})\\=1-2^{-s}+ \sum_{n=2}^\infty F_k(n) (n^{-s}-(n+1)^{-s})=1-2^{-s} -\sum_{n=2}^\infty \sum_{m=0}^{k+1} c_{m,k} n^m n^{-s} \sum_{l=1}^\infty {-s \choose l} n^{-l}$$ $$ =1-2^{-s} -\sum_{m=0}^{k+1} c_{m,k} \sum_{l=1}^\infty {-s \choose l} (\zeta(s+l-m)-1) \tag{1}$$

Also $\zeta(s) = \sum_{n=1}^\infty \int_n^\infty s x^{-s-1}dx =s \int_1^\infty \lfloor x \rfloor x^{-s-1}dx =\frac{s}{s-1}+s \int_1^\infty (\lfloor x \rfloor -x)x^{-s-1}dx$ thus $\lim_{s \to 1} (s-1)\zeta(s) = 1$ and together with $(1)$ it shows $(s-1)\zeta(s)$ is analytic everywhere.

Therefore letting $s \to 0$ in $(1)$, noting $\lim_{s \to 0} {-s \choose l}\zeta(s+l-m) = 0$ for $l-m \ne 1$ $$\zeta(0-k) =-\sum_{m=0}^{k+1} c_{m,k}\prod_{j=1}^{m} \frac{-0-j}{j+1}= \int_{-1}^0 F_k(t)dt $$

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  • $\begingroup$ This seems clear to me, except one step. How do you get from summing $n^kn^{-s}$ to summing $F_k(n)(n^{-s}-(n+1)^{-s})$? $\endgroup$ – Michal Dvořák Dec 28 '17 at 0:28
  • $\begingroup$ @MichalDvořák Summation by parts $\endgroup$ – reuns Dec 28 '17 at 0:32
  • $\begingroup$ Wait, that correcture you just did doesn't appeal to me, why would you take out the $n=1$ term from the sum. Then you can't get the riemann zeta function out. It seems correct because $1-2^{-s}$ is zero as $s$ goes to zero. Also would you please also explain me how did you come from the sum of the product to the integral? (The very last line). $\endgroup$ – Michal Dvořák Dec 28 '17 at 10:43
  • $\begingroup$ @MichalDvořák It is to make things absolutely convergent, $\zeta(s+l-m)-1$ is exponentially decreasing as $ l \to \infty$. And $\prod_{j=1}^{m} \frac{-0-j}{j+1} = \frac{(-1)^m m!}{(m+1)!}$. $\endgroup$ – reuns Dec 28 '17 at 10:49
  • $\begingroup$ Great, may i ask one more just to be sure, $c_{m,k}$ is the coefficient of $m$-th term of the Faulhauber's polynmial for $k$-th power, is that correct? $\endgroup$ – Michal Dvořák Dec 28 '17 at 10:58
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Note that $F_{k+1}'(x)-F_{k+1}'(0)=(k+1)F_k(x)$. So $$(k+1)\int_{-1}^0 F_k(x)\,dx=-F_{k+1}'(0)+F_{k+1}(0)-F_{k+1}(-1) =-F_{k+1}'(0)+0^k=-F_{k+1}'(0).$$ But $F_{k+1}'(0)=(-1)^{k+1} B_{k+1}$ so we get your formula.

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  • $\begingroup$ It's obvious that $\int{F_k(n)dn}=\frac{F_{k+1}}{k+1}$ But what the term $-F_{k+1}(0)'$ has to do there. And what about the $F_{k+1}(-1)$. By the notation i introduced, one can not sum first $-1$ positive integers, that somehow doesn't make sense. Would you please explain more thoroughly what you actually did? $\endgroup$ – Michal Dvořák Dec 27 '17 at 22:34
  • $\begingroup$ Note that $F_k$ is characterised as the unique polynomial with $F_k(x)-F_k(x-1)=x^k$ and $F_k(0)=0$. @MichalDvořák $\endgroup$ – Lord Shark the Unknown Dec 28 '17 at 3:28
  • $\begingroup$ Okay, thats with the $F_k(-1)$ term, but i still dont understand, where did that $-F'_{k+1}(0)$ term appear. and why it should equal $(-1)^{k+1}B_{k+1}$ It appears literally from nowhere and it should equal zero. $\endgroup$ – Michal Dvořák Dec 28 '17 at 9:30

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