Convergence a.s from a sub-sequence Let $(X_n)_n$ be a sequence of independent random variables. Let $Y_n=\frac{1}{n}\sum_{k=1}^nX_k.$
Prove that if $(Y_n)_n$ converges in probability to $0$ and if $Y_{2^n}$ converges a.s to $0$ then $Y_n$ converges a.s to $0$.
$Y_n$ converges a.s if $\forall \epsilon>0,\lim_n P(\sup_{k \geq n}|Y_k|>\epsilon)=0,$ any ideas how to use the above facts to prove the result?

I think I have it.  If this is the proof they were looking for, it sure seems difficult.
Denote $S_n = \sum_{k=1}^n X_k$.  Write
$ Y_n = \tfrac1n S_{2^l} + Z_n $, where $2^l \le n < 2^{l+1}$, and
$$ Z_n = \frac1n \sum_{k=2^l}^n X_k .$$
Since $|\frac1n S_{2^l}| \le |Y_{2^l}|$, $\frac1n S_{2^l} \to 0$ a.s.  Now by Etemadi's inequality: https://en.wikipedia.org/wiki/Etemadi%27s_inequality
$$ P(\sup_{2^l \le n < 2^{l+1}} |Z_n| > \epsilon ) \le 3 \sup_{2^l \le n < 2^{l+1}}  P(|Z_n| > \epsilon / 3 ) = 3 P(|Z_{n_l}| > \epsilon / 3 ) $$
for some $2^l \le n_l < 2^{l+1}$.
Let $V_l = \frac1{n_l}(S_{2^{l+1}}-S_{n_l})$.  Using the fact that $Y_n \to 0$ in probability, we can show that $V_l \to 0$ in probability.  Hence there exists $L_1$ such that for $l \ge L_1$, we have $P(|V_l| > \epsilon/6) \le 1/2$.  Now
$$ \{|Z_{n_l} + V_l| > \epsilon/6 \} \supset \{ |V_l| \le  \epsilon / 6\} \cap \{|Z_{n_l} |>\epsilon/3\} .$$
Taking probabilities, and noting that $Z_{n_l}$ and $V_l$ are independent, we obtain
$$ P(|Z_{n_l} + V_l| > \epsilon/6 ) \ge \tfrac12 P(|Z_{n_l} |>\epsilon/3) .$$
But
$$ |Z_{n_l} + V_l| = \frac{2^l}{n_l} |2 Y_{2^{l+1}} - Y_{2^l}| .$$
So
$$ P(|Z_{n_l} |>\epsilon/3) \le 2 P(|2 Y_{2^{l+1}} - Y_{2^l}| > \epsilon/12).$$
Let
$$ W_l = \sup_{2^l \le n < 2^{l+1}} |Z_n| .$$
Note that the $W_l$ are independent.  So by the Borel-Cantelli lemma
$$ P(\sup_{n>2^L} |Z_n| > \epsilon ) = P(\sup_{l>L} W_l > \epsilon) $$
converges to $0$ if and only if $\sum_{l=1}^\infty P(W_l > \epsilon)$ converges, which in turn happens if $\sum_{l=1}^\infty P(|2Y_{2^{l+1}} - Y_{2^l} > \epsilon/12)$ converges.  And because the random variables $2 Y_{2^{l+1}} - Y_{2^l}$ are independent, using the Borel-Cantelli lemma again, the last sum converges if and only if
$$ P(\sup_{l \ge L} |2 Y_{2^{l+1}} - Y_{2^l}| > \epsilon/12) \to 0 .$$
But $2 Y_{2^{l+1}} - Y_{2^l} \to 0$ a.s.  So Egorov's Theorem completes the proof that $Z_n \to 0$ a.s., and from this we see that $Y_n \to 0$ a.s.
