Is there any procedure to find an upper bound of the following expression?
$$\frac{1}{1-\beta u}-\sum_{n=0}^{\infty}\frac{e^{-u}u^n}{n!(1-\beta(n+1))}$$ Here $u,\beta\in\mathbb{R},\ u>1,\ 0\le\beta<1,\ \beta u<1$ and $\beta$ is assumed to be not of the form $1/k$ for any $k\in\mathbb{Z^+}$.
Note that $\beta=0$ implies that the expression is $0$.
Edit: Specifically I want to ask the question
Given any $\epsilon>0$ can there be a $\beta\in (0,1)$ such that I can make $$\left|\frac{1}{1-\beta u}-\sum_{n=0}^{\infty}\frac{e^{-u}u^n}{n!(1-\beta(n+1))}\right|<\epsilon$$ Also, it will be very useful for me if anything can be said about how the choice of $\beta$ depends upon the choice of $\epsilon$. Thanks in advance.
Further Edit: I figured out that the answer to my question is negative. Here is my argument below. If anyone finds any flaw please kindly indicate it. that will be a lot of help. Thanks.
$n\ge 0,\ \beta>0\Rightarrow \frac{1}{1-\beta(n+1)}\ge \frac{1}{1-\beta}$. So, $$\frac{1}{1-\beta u}-\sum_{n=0}^\infty \frac{e^{-u}u^n}{n!(1-\beta(n+1))}\le \frac{1}{1-\beta u}-\frac{1}{1-\beta}=\frac{\beta(u-1)}{(1-\beta u)(1-\beta)}$$ So, if I choose $$\frac{\beta(u-1)}{(1-\beta u)(1-\beta)}=\epsilon\Rightarrow \beta^2 u\epsilon- 2u\beta+1=0\Rightarrow \beta=\frac{u\pm\sqrt{u^2-u\epsilon}}{u\epsilon}$$ So, $u\ge\epsilon$ and also, $\beta<\frac{1}{u}\Rightarrow \sqrt{u^2-u\epsilon}<\epsilon-u\Rightarrow \epsilon>u$. So, this inequality is not possible for any $\beta<\frac{1}{u}$.