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 $k+\epsilon <|\lambda| < k+1-\epsilon$. Assume that the so defined function on $\{\lambda\in\mathbb{C}, ||\lambda|-n|>\epsilon, \forall n\in\mathbb{Z}\}$ is bounded on every strip $|\Im\lambda|<c$ intersected with its domain. Finally assume that $k+1-\epsilon$ is the convergence radius of the $k$-th series.
I would like to show that $a_n^k\to 0$ for all $n>0$. Intuitively I'm convinced however, without the constraint on the convergence radius this wouldn't be true since $\sin(\lambda)$ would be a counter example.
Here my very sketchy idea how this could go (which doesn't quite work). I started using that the convergence radius can be computed using Cauchy-Hadamard to be $$ \limsup_{n\mapsto\infty} \sqrt[n]{|a_n^k|} = \frac{1}{k+1-\epsilon} $$ which is monotonely decreasing in $k$. Hence $\sqrt[n]{|a_n^k|}$ is essentially decreasing in $k$ which means that $a_n^k$ are essentially negative. If I can show that allmost all coefficients are negative somehow then I obtain that each $a_n^k\lambda^n$ is bounded except if I get cancellation from negative powers. Maybe I can bound the cancellation using a similar argument on the negative part and the lower bounds. For this I could use (what I didn't mention) that $f(\lambda^{-1})$ fullfils the same conditions. Hence essentially all coefficients are negative. I just need to understand what the essentially means in all this and how to use it. Any idea is very wellcome.
This question is a second version of my former question same question but with uniform boundedness.