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?


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

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 $\endgroup$ – Angina Seng 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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