Summability of random series given an expectation condition Let $(Y_n)$ be a collection of i.i.d nonnegative random variables with $E[\log^+Y_1]$ finite, where $\log^+ X=\max(0,\log(X))$. How does one show that this is equivalent to the fact that
$
\sum\limits_{k=0}^\infty \frac{Y_k}{c^k}
$ is finite a.s. for any $c>1$?
 A: I will assume $Y_n$ are positive a.s. such that $\log Y_1$ are well-defined. Clearly, we have
$$ (\log a) \mathbb{P}(a < Y_n \leq b) \leq \int_{\{a < Y_n \leq b\}} \log Y_n \; d\mathbb{P} \leq (\log b) \mathbb{P}(a < Y_n\leq b).$$
In particular for $(a, b) = (c^k, c^{k+1})$ for $k = 0, 1, 2, \cdots$,
$$ k (\log c) \mathbb{P}\left(c^{k} < Y_n \leq c^{k+1}\right) \leq \int_{\{c^k < Y_n \leq c^{k+1}\}} \log^{+} Y_n \; d\mathbb{P} \leq (k+1)(\log c) \mathbb{P}\left(c^{k} < Y_n \leq c^{k+1}\right). $$
Thus if we let
$$ S(c) = (\log c) \sum_{k=1}^{\infty} k \mathbb{P}\left(c^{k} < Y_n \leq c^{k+1}\right),$$
then summing the inequality above gives
$$ S(c) \leq \mathbb{E}[\log^{+} Y_n] \leq S(c) + \log c. $$
Since $Y_n$ are i.i.d., $S$ does not depend on $n$. Thus this inequality implies that $\mathbb{E}[\log^{+} Y_1] < \infty$ if and only if $S(c) < \infty$. Here, we observe that the finiteness of $S(c)$ also does not depends on the choice of $c > 1$.
Now rearranging,
$$\begin{align*}
S(c)
&= (\log c) \sum_{k=1}^{\infty} \sum_{j=1}^{k} \mathbb{P}\left(c^{k} < Y_1 \leq c^{k+1}\right) \\
&= (\log c) \sum_{j=1}^{\infty} \sum_{k=j}^{\infty} \mathbb{P}\left(c^{k} < Y_1 \leq c^{k+1}\right) \\
&= (\log c) \sum_{j=1}^{\infty} \mathbb{P}\left( c^j < Y_1 \right) \\
&= (\log c) \sum_{j=1}^{\infty} \mathbb{P}\left( c^j < Y_j \right)
\end{align*}$$
Now we prove the prescribed equivalence.
$(\Longrightarrow)$ : By our previous observation, $\mathbb{E}[\log^{+} Y_n] < \infty$ if and only if $S(b) < \infty$ for $1<b<c$. The first Borel-Cantelli lemma says that $S(b) < \infty $ implies
$$ \mathbb{P} \left( b^{j} < Y_j \text{ for infinitely many } j \right) = 0. $$
Thus $Y_j / c^j < (b/c)^{j}$ except for finitely many exceptions a.s., and hence the finiteness of $\sum_{j} Y_j / c^j$ a.s..
$(\Longleftarrow)$ : We prove the contraposition. Assume $\mathbb{E}[\log^{+} Y_n] = \infty$. This is equivalent to say that $S(c) = \infty$. Since $Y_j$ are i.i.d., the events $\{ c^{j} < Y_j \}$ are mutually independent, hence we can apply the second Borel-Cantelli lemma to obtain
$$ \mathbb{P} \left( c^{j} < Y_j \text{ for infinitely many } j \right) = 1. $$
This proves that $\sum_{j} Y_j / c^j$ is not a.s. finite (in fact, infinite a.s.). ////
