Let $(X\_n)\_{n\in\mathbb{N}}$ be a sequence of i.i.d. real-random variables. Let further $0<c<1$. I'm interested in the asymptotic properties of
$$
\sum\_{k=1}^n c^k X\_k.
$$
I can prove that this converges a.s. for $n\to\infty$ iff $\mathbb{E}(\max(0,\log(|X\_1|)<\infty$.
To be more specific:


* Are there known conditions for $\limsup\limits\_{n\to\infty}\sum\_{k=1}^n c^k X\_k=\infty$ a.s.?
* As far as I know, the asymptotic behaviour of $c^n\sum\_{k=1}^nX\_k$ is fairly well understood (c.f. ["A note on Fellers strong law of large numbers" by Chow, and Zhang (1986)](https://projecteuclid.org/journals/annals-of-probability/volume-14/issue-3/A-Note-on-Fellers-Strong-Law-of-Large-Numbers/10.1214/aop/1176992464.full)). Is there a known connection (or differences) between the asymptotic behaviour of $c^n\sum\_{k=1}^nX\_k$ and $\sum\_{k=1}^n c^k X\_k$? E.g. is it possible that $\lim\limits\_{n\to\infty}c^n\sum\_{k=1}^nX\_k=\infty$ a.s., but $\lim\limits\_{n\to\infty}\sum\_{k=1}^n c^k X\_k<\infty$ a.s.? I can only proof that if $\lim\limits\_{n\to\infty}c^n\sum\_{k=1}^nX\_k=\infty \text{ a.s.} \Rightarrow \lim\limits\_{n\to\infty}\sum\_{k=1}^n c^k X\_k$ diverges a.s.
* I'm having trouble constructing examples, where $\lim\limits\_{n\to\infty}\sum\_{k=1}^n c^k X\_k=\infty$ a.s. holds. Does anybody know a way to construct such examples?


I would be very grateful for any advice or reference!



Asymptotic properties of weighted random walks / infinite convolutions of random variables
Let $(X\_n)\_{n\in\mathbb{N}}$ be i.i.d. real random variables and let $0<c<1$.  

Then the following are equivalent:


(a) There exists $r>0$ such that $P(|X\_k|>e^{rk} \; \, \text{infinitely often} )=0$.


(b) $\mathbb{E}(\max(0,\log(|X\_1|)<\infty$.


(c) For all $r>0$, we have $P(|X\_k|>e^{rk} \; \, \text{infinitely often} )=0$.


(d) The series $ \sum\_{k=1}^n c^k X\_k $ converges a.s.


(e) $\lim \_n c^n S\_n =0 \;$ a.s., where $S\_n:=\sum\_{k=1}^n X\_k$.


**Proof:** For a random variable $Y \ge 0$, we have
$$\sum\_{k=1}^\infty {\bf 1}\_{\{Y \ge k\}} \le Y \le \sum\_{k=0}^\infty {\bf 1}\_{\{Y \ge k\}} \,,$$
so taking expectations gives the well-known inequalities
$$\sum\_{k=1}^\infty P(Y \ge k) \le \mathbb{E}(Y) \le \sum\_{k=0}^\infty P(Y \ge k) \,.$$
Thus by the Borel-Cantelli lemma, if $Y\_k \ge 0$ are i.i.d. and $r>0$, then
$$\mathbb{E}(Y\_1) <\infty \Longleftrightarrow \mathbb{E}( Y\_1/r) <\infty \Longleftrightarrow $$ $$ \sum\_k P(Y\_k \ge {rk}) <\infty \Longleftrightarrow P(Y\_k \ge {rk} \; \, \text{infinitely often} )=0 \, . \tag{\*}$$


Applying this to $Y\_k=\max(0,\log(|X\_k|)$ shows that
$ \;(a) \Longrightarrow(b) \Longrightarrow (c)$. Obviously $(c) \Longrightarrow (a)$.


Clearly $\; (c) \Longrightarrow (d) \Longrightarrow (a)$ and
$\; (c) \Longrightarrow (e)$.


Finally, suppose (e) holds. Then
$c^n X\_n =(c^n S\_n) - c(c^{n-1}S\_{n-1}) \to 0$ a.s.,
and (a) follows.


QED


