Let $(X\_n)\_{n\in\mathbb{N}}$ be a sequence of strictly positive and identically distributed random variables and let $\beta\le 1$. I am trying to prove that 



> 
> $$
> 0<\lim\_{\beta\rightarrow 1}(1-\beta)\sum\_{n=0}^{\infty}\beta^nX\_n <\infty.
> $$
> 
> 
> 


I have already shown that the limit is finite if $\mathbb{E}\log\_+(X\_n))<\infty$, but I am having trouble showing the nonzero part. Does anybody have a reference or idea on how to prove this result?


Thank you in advance!



Divergence rate of geometric sum of random variables
$\newcommand{\be}{\beta}
\newcommand{\E}{\operatorname{\mathsf E}} $ 


Note that $\mu:=\E X\_1\in(0,\infty]$. 
Suppose first that $\mu<\infty$. 
Then, for $\be\uparrow1$,
\begin{align\*}
 (1-\be)\sum\_{n=0}^{\infty}\be^n X\_n
 &=(1-\be)^2\sum\_{n=0}^\infty X\_n\sum\_{j=n}^\infty\be^j \\ 
 & =(1-\be)^2\sum\_{j=0}^\infty \be^j\sum\_{n=0}^j X\_n \\ 
 & \sim(1-\be)^2\sum\_{j=0}^\infty \be^j (j+1)\mu =\mu; 
\end{align\*}
here we used the strong law of large numbers and the fact that $(1-\be)^2\sum\_{j=0}^N \be^j\sum\_{n=0}^j X\_n\to0$ for each natural $N$.  

So, 
\begin{equation}
 \lim\_{\be\uparrow1}(1-\be)\sum\_{n=0}^{\infty}\be^n X\_n=\mu \tag{1}
\end{equation}
if $\mu<\infty$. By using the truncation $X\_n1\_{X\_n\le C}$, it is now easy to see that (1) holds for $\mu=\infty$ as well. 


(In particular, it follows that the limit does not have to be finite if you only have $\E\log\_+(X\_n)<\infty$.)


