Divergence rate of geometric sum of random variables 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!
 A: $\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 $\E\log_+(X_n)<\infty$.)
A: Write 
\begin{align*}
S_\beta &=(1-\beta)\big[X_0+X_1+X_2+\dots\big]
\\
&=(1-\beta)\big[(1-\beta)X_0 + (\beta-\beta^2)(X_0+X_1) + (\beta^2-\beta^3)(X_0+X_1+X_2) + \dots\big]
\\
&=(1-\beta)^2\big[X_0 + \beta(X_0+X_1) + \beta^2 (X_0+X_1+X_2)+ \dots\big].
\end{align*}
Suppose $X_n$ has finite mean $\mu$. Let $\epsilon>0$. By the Strong Law of Large Numbers, with probability $1$, $X_0+X_2+\dots+X_{n-1}>n(\mu-\epsilon)$ for all large enough $n$. In that case, 
\begin{align*}
\liminf_{\beta\uparrow 1} S_\beta 
&\geq \liminf_{\beta\uparrow 1} (\mu-\epsilon)(1-\beta)^2[1+2\beta+3\beta^2+\dots]\\
&=\mu-\epsilon.
\end{align*}
In a similar way, with probability $1$, $\liminf_{\beta\uparrow1} S_\beta\leq \mu+\epsilon$. So you have that with probability $1$, the limit $\lim_{\beta\uparrow1} S_\beta$ exists and equals $\mu$. 
In the case where the mean is infinite, the same argument gives that with probability $1$, $S_{\beta} \to \infty$ as $\beta\uparrow1$. This seems to contradict your statement that the limit is finite whenever $\mathbb{E} \log_+ X_n<\infty$.
