Let $\{c\_n\}$ be a descending sequence of positive real numbers, and let $\{X\_i\}$ be a sequence of i.i.d. random variables.


Are the following statements equivalent?


1. $\operatorname{E}(X\_1^2) < \infty$ and $\operatorname{E}(X\_1) = 0$ implies $\lim \limits\_{n \to \infty} c\_n \sum \limits\_{i = 1}^n X\_i \xrightarrow{a.s.} 0$.
2. $\sum \limits\_{i = 1}^\infty c\_i^2 < \infty$ .


Thank you for any replies or comments. 



strong law of large number
The [Law of the Iterated Logarithm](http://en.wikipedia.org/wiki/Law_of_the_iterated_logarithm) tells you 
$$
\limsup\_{n\to\infty}\left|\frac{S\_n}{\sqrt {n\log\log n}}\right|=\sigma\sqrt 2\text{ with probability 1,}
$$
where $S\_n:=X\_1+\ldots+X\_n$, and $\sigma^2=\mathbb EX\_1^2$. Hence if 
$\sqrt{n\log\log n}\cdot c\_n\to 0$, one has $c\_nS\_n\to 0$ with probability 1. 


This is not quite a necessary and sufficient condition, because you can construct decreasing sequences $(c\_n)$ that *most of the the time* are of smaller order than $1/\sqrt{n\log\log n}$, but occasionally are as big as $1/\sqrt{n\log\log n}$, for which $c\_nS\_n$ still goes to 0 almost surely, but if you want $c\_n$ to be some kind of regularly decaying sequence, then this is exactly the right condition.


