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We know the Dirichlet $\eta$-function is defined as the analytic continuation of $$\eta(s) = \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^s} \quad \Re(s)>0$$ I find an identity for the values of this function at negative integers: $$\eta(-n)+ \sum_{k=0}^n {n\choose k}(-1)^k \eta(k-n) =0.$$ where $1<n\in\mathbb{Z}$.

I found this using some "computation" on divergent series. But this wrong computation cannot serve as a proof.

I think this identity should be true, though I haven't found a simple proof.

Do you know anything about this identity? If it's true, how to prove it? Is there any generalization (for arbitrary $s$ rather than $-n$) of this identity? Could you provide me some reference?

Riemann
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Fang Hung-chien
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1 Answers1

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1) Bernoulli numbers, $(n=0,1,2,...; B^n:=B_n)$, (DLMF = http://dlmf.nist.gov/24)

\begin{align*} & B = [1, -1/2, 1/6, 0, -1/30, 0, ...] \\ &\iff e^{Bx} = x/(e^x-1) \\ &\iff e^{(B+1)x} - e^{Bx} = x \\ &\iff (B+1)^{n+1} - B^{n+1} = \delta_{n,0} \qquad{(\rm DLMF\; 24.4.1,}\,x=0) \\ &\iff \cosh(Bx) = (x/2)\coth(x/2) \quad\&\quad \sinh(Bx) = -x/2 \end{align*}

2) Bernoulli polynomials at $x=1/2$, $(n=0,1,2,...; B^n:=B_n)$ \begin{align*} & e^{(B+1/2)x} = e^{Bx} (e^{x/2}+1) - e^{Bx} = 2 (x/2)/(e^{x/2}-1) - x/(e^x-1) \\ &\iff (B+1/2)^n = (2^{1-n}-1) B_n \qquad{(\rm DLMF\; 24.4.27}) \\ &\iff (2B+1)^n = (2-2^n) B_n \end{align*}

3) Dirichlet's $\eta(s)$ function and Riemann's $\zeta(s)$ function, $(n=0,1,2,...)$

\begin{align*} \eta(-n) &= (1-2^{n+1}) \zeta(-n) \\ \zeta(-n) &= (-1)^n B_{n+1}/(n+1)\qquad({\rm DLMF\; 25.6.3, avec}\;n=0^*) \end{align*} *Jonathan Sondow, Analytic continuation of Riemann's zeta function and values at negative integers via Euler's transformation of series http://www.ams.org/journals/proc/1994-120-02/S0002-9939-1994-1172954-7

4) Récurrence pour $\eta(-n), (n=0,1,2,...; B^n:=B_n)$ \begin{align*} &\;(-1)^n (n+1) \left[ \;\eta(-n) + \sum_{k=0}^n {n\choose k} (-1)^{n-k} \eta(-k) \;\right] \\ &= (1-2^{n+1}) B_{n+1} + \sum_{k=0}^n {n\choose k} (n+1)/(k+1) (1-2^{k+1}) B_{k+1} \\ &= (1-2^{n+1}) B_{n+1} + \sum_{k=0}^n {n+1\choose k+1} (1-2^{k+1}) B_{k+1} \\ &= (1-2^{n+1}) B_{n+1} + \sum_{k=0}^{n+1} {n+1\choose k} (1-2^k) B_k \\ &= \left[(2-2^{n+1}) B_{n+1} - (2B+1)^{n+1}\right] + \left[(B+1)^{n+1} - B^{n+1}\right] \\ &= 0 + \delta_{n,0} \\ &= (-1)^n (n+1) \left[\; \delta_{n,0} \;\right] \qquad{\rm CQFD} \end{align*} Jacques Gélinas

Scientifica
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Jacques
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