Let $a_n=\frac{\Gamma(n+h)}{\Gamma(n+1)\Gamma(h)}$ for $0<h<1$ and $n\geq0$ and define $b_n$ such that $\sum_{0}b_n=b<\infty$ and $b\neq0$. Now define $$c_n=\sum_{m=0}^nb_ma_{n-m}.$$ Can we say that $$\lim_{n\to \infty}n^{1-h}c_n=\frac{b}{\Gamma(h)}$$ if and only if $$\lim_{n\to \infty}n^{1-h}b_n=0.$$
I will be thankful for a counter example.
To prove this I tried using dominated convergence theorem, but that does not work since I don't know if $b_n$ is tight. I appreciate any hints or ideas to approach this.