Equivalence of $\frac{S_n}{n}\xrightarrow{a.s.}0$ I'm trying to prove the next equivalence:
Let $\{X_n\}_{n\in\mathbb{N}}$ be a sequence of independent r.v. and $S_n=\sum_{i=1}^{n}X_i.$ 
Then $\frac{S_n}{n}\xrightarrow{a.s.}0$ if and only if the two following conditions holds:
a) $\frac{S_n}{n}\xrightarrow{P}0,$
b)$\frac{S_{2^n}}{2^n}\xrightarrow{a.s.}0.$
If $\frac{S_n}{n}\xrightarrow{a.s.}0$ then we have immediatly a) because convergence a.s. implies in probability and every subsequence converges to $0,$ so b) is satisfied.
For the other direction I am having problems. I was trying to use that every subsequence of $\frac{S_n}{n}$ has some subsequence which converges a.s; then mixed each term of such subsequence with b), but this cannot ensure the desired result.
Any kind of help is thanked in advanced.
 A: 
Lemma. Let $X_1,X_2,\cdots$ be independent and $S_{m,n}=X_{m+1}+\cdots+X_n$. Then for any $a>0$, we have
  $$P\left(\max_{m<j\leq n} |S_{n,j}|>2a\right)\min_{m<k\leq n}P(|S_{k,n}|\leq a)\leq P(|S_{m,n}|>a).$$

Proof. Set $S$ and $T$ by 
\begin{align*}
S&=\inf\{j>m:|S_{m,j}|>2a\text{ and }|S_{j,n}|\leq a\},\\
T&=\inf\{j>m:|S_{m,j}|>2a\}
\end{align*}
where $\inf\varnothing=\infty$. Then by the triangle inequality,
\begin{align*}
\{S\leq n\}&=\{|S_{m,j}|>2a\text{ and }|S_{j,n}|\leq a\text{ for some }j\in(m,n]\}\\&\subset\{|S_{m,n}|>a\}.
\end{align*}
We also find that 
$$\{T=k\}\cap\{|S_{k,n}|\leq a\}\subset\{S=k\}.$$
Obeserve that the event $\{T=k\}$ only depends on $X_{m+1},\cdots,X_k$ while the event $\{|S_{k,n}|\leq a\}$ only depends on $X_{k+1},\cdots,X_n$. So these events are independent. Then 
\begin{align*}
P(|S_{m,n}|>a)&\geq P(S\leq n)=\sum_{k=m+1}^n P(S=k)\\&\geq P(T=k)P(|S_{k,n}|\leq a)\geq P(T\leq n)\min_{m<k\leq n}P(|S_{k,n}|\leq a).
\end{align*}
Therefore the desired inequality follows from the equality $$P(T\leq n)=P\left(\max_{m<j\leq n} |S_{n,j}|>2a\right).$$

Corollary. Let $X_1,X_2,\cdots$ be independent and $S_{n}=X_{1}+\cdots+X_n$. If $S_n/n\to0$ in probability and $S_{2^n}/2^n\to0$ a.s., then $S_n/n\to 0$ a.s.

Proof. Let $S_{m,n}=S_n-S_m$ and $\epsilon>0$. Since $|S_{m,n}|\leq|S_m|+|S_n|$ and $S_n/n\to0$ in probability, we have 
$$\min_{2^{n-1}<m\leq 2^n}P(|S_{m,2^n}|\leq \epsilon2^n)\to 1\ \ \text{as }n\to\infty.$$
Hence for $n$ large enough,
$$P\left(\max_{2^{n-1}<j\leq 2^n} |S_{2^{n-1},j}|>2\epsilon2^n\right)\leq 2 P(|S_{2^{n-1},2^n}|>\epsilon2^n).$$
Since $S_{2^n}/2^n\to0$ a.s., the events on the right hand side are independent and only occur finitely often, so Borel-Cantelli lemma implies that their probabilities are summable and so are the probabilities of events on the left hand side. Another application of Borel-Cantelli lemma implies that the events on the left hand side only occur finitely often. Finally, using that for $2^{n-1}<j\leq2^n$,
$$\frac{|S_j|}j\leq\frac{|S_j|}{2^{n-1}}=\frac{|S_{2^{n-1},j}+S_{2^{n-1}}|}{2^{n-1}}\leq2\frac{|S_{2^{n-1},j}|}{2^n}+\frac{|S_{2^{n-1}}|}{2^{n-1}}$$
and $S_{2^n}/2^n\to0$ a.s. gives the desired result.
