An identity about Dirichlet $\eta$ Function 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? 
 A: 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
