Let $ \{c_n\} $ be a descending positive real sequence.
Let $ X_1,X_2,\cdots $ be a sequence of i.i.d random variables.
Is the following equivalent?
($1$) For any i.i.d sequence $ X_1,X_2,\cdots $ such that$ \operatorname{E}(X_1^2)<\infty, \operatorname{E}(X_1)=0$ . $ \lim\limits_{n\to\infty}c_n\sum\limits_{i=1}^n X_i\to 0 $ a.s
($2$) $ \sum\limits_{n}c_n^2<\infty $
Thanks for your reply.
On
Recall the law of the iterated logarithm:
Let $(X_n)_{n \in \mathbb{N}}$ be a sequence of iid random variables such that $\mathbb{E}(X_1^2)=:\sigma^2<\infty$ and $\mathbb{E}X_n=0$. Then, $$\limsup_{n \to \infty} \left|\frac{X_1+\ldots+X_n}{\sqrt{2n \log \log n}} \right| = \sigma \quad \text{a.s.}$$
This means in particular that $(1)$ does not imply $(2)$. In fact, if we choose
$$c_n := \frac{1}{\sqrt{\log n}} \cdot \frac{1}{\sqrt{n \log \log n}},$$
then it follows from the law of iterated logarithm that
$$c_n \cdot \sum_{i=1}^n X_i = \frac{1}{\sqrt{\log n}} \frac{\sum_{i=1}^n X_i}{\sqrt{n \log \log n}} \to 0.$$
On the other hand,
$$\sum_{n=1}^{\infty} c_n^2 = \sum_{n=1}^{\infty} \frac{1}{n \cdot \log n \cdot \log \log n} = \infty.$$
Using a very similar argumentation, one can show that $(2)$ implies $(1)$.
i see to use the law of the iterated logarithm that tell us: $\limsup_{n\to\infty}\left|\frac{S_n}{\sqrt {n\log\log n}}\right|=\sigma\sqrt 2\text{ with probability 1,}$
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 we 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\to 0 $still goes to 0 almost surely , but if we want$ c_n$ to be some kind of regularly decaying sequence, then this is exactly the right condition.
i looked this book :The Theorem 2.5.7, p. 71 :http://www.math.duke.edu/~rtd/PTE/pte.html