Let $X_{i}$ be independent random variables,$S_n=\sum_1^n X_i.\ $if $\lim_{n\to \infty} \frac{S_n}{n}\to0 $ in Pr and $\lim_{n\to \infty} \frac{S_{2^n}}{2^n}\to0 \ a.s.$, prove $\lim_{n\to \infty} \frac{S_n}{n}\to0\ a.s.$ .
This problem is from Chung's book,it's also appear in Durrett's book with additional condition that $X_i\ $are identically distributed. In durrett's book,there is a hint:when $X_i\ $are independent,$P(\max_{m<j\leq n}|S_{m,j}|>2\epsilon)\cdot\min_{m<k\leq n} P(|S_{k,n}|\leq \epsilon)\leq P(|S_{m,n}|>\epsilon)$,where $S_{m,n}=S_n-S_m$.
With the convergence in Pr condition,we get $\min_{2^{n-1}<j\leq 2^{n}} P(|S_{j,2^n}|\leq \epsilon 2^n)\to 1$.As a result,using B-C lemma,we get $\sum P(\max_{2^{n-1}<k\leq 2^{n}} |S_{2^{n-1},j}|>\epsilon 2^{n+1})<\infty$ .It seems we don't use the identically distributed condition in above discussion,so the condition is unnecessary?
Is there any other more natural solution?
In order to deduce that $$\sum P\left(\max_{2^{n-1}<k\leq 2^{n}} \left\lvert S_{2^{n-1},j}\right\rvert >\epsilon 2^n\right)<\infty$$ from the lemma, you also need that $\sum_{n\geq 1}\mathbb P(S_{2^{n-1},2^n})>2^n\varepsilon)$ to be finite. As $(S_{2^{n-1},2^n}/2^n)_{n\geqslant 1}$ is independent, this is a consequence of the fact that $S_{2^{n-1},2^n}/2^n\to 0$ almost surely and for this step, it is indeed not required that the $X_i$ have the same distribution.