Let $a_n^k$ be real coefficients, such that for each $n\geq 0$ $(a_n^k)_{k\geq0}$ is monotonely increasing and such that $$ \sum_{n\in\mathbb{Z}} a_n^k \lambda^n $$ is a holomorphic function on $k+\epsilon <|\lambda| < k+1-\epsilon$. Finally assume that the so defined function on $\{\lambda\in\mathbb{C}, ||\lambda|-n|>\epsilon, \forall n\in\mathbb{Z}\}$ is bounded.
I would like to show that $a_n^k\to 0$ for all $n>0$. Intuitively I'm convinced however, if I take a sequence $N<\lambda_N<N+1$ and I want to use the boundedness, I can't be sure that there is no cancelation from the negative coefficients getting bigger.
Is there an elegant prove? (If at all, a counter example would blow me away but would be even more important to know)
The integral formula for Laurent coefficients gives that for any path $\gamma_k$ around zero such that $k+\epsilon<|\gamma_k | <k+1-\epsilon$ one has $$ a_n^k = \frac{1}{2\pi i}\int_{\gamma_k} \frac{f(z)}{z^{n+1}} \mathrm{d}z. $$ By the boundedness of $f$ on then sees that $a_n^k\to 0$ for all $n\geq1$ as $k\mapsto\infty$ while $a_0^k$ must be bounded and hence cenverges.