Let $(X_n)$ be a sequence of independent real random variables and $(\tau_n)$ be a null sequence of real numbers. Prove that either each of the sequences $(X_n)$ and $ \big (\tau_n\sum_{i=1}^n X_i \big)$ converges with probability $1$ or each diverges with probability $1$.
Let $A$ be the event that $(X_n)$ converges and $B$ be the event that $ \big (\tau_n\sum_{i=1}^n X_i \big)$ converges. I was able to show that both $A$ and $B$ are tail events, so that by Kolmogorov's $0$-$1$ law we have $P(A)=0 \text{ or }1$ and $P(B)=0 \text{ or }1$. But how can I rule out the case $P(A)\neq P(B)$?
This is a preliminary exercise to show that if $(X_n)$ is independent and obeys the strong law of large numbers then
$$ \sum_{n=1}^{\infty} P\{n^{-1} |X_n|\geq \epsilon\} <\infty$$
for all $\epsilon>0$.
Thanks a lot for your help.
If the strong law of large numbers holds then there is a real number $\mu$ such that $\frac{X_1+...+X_n}{n}\to\mu$ almost surely. Now, at every point where it happens we have:
$\frac{X_n}{n}=\frac{X_1+...+X_n}{n}-\frac{X_1+...+X_{n-1}}{n}=\frac{X_1+...+X_n}{n}-\frac{X_1+...+X_{n-1}}{n-1}\frac{n-1}{n}\to\mu-\mu=0$
And so $\frac{X_n}{n}\to 0$ almost surely.
Now let $\epsilon>0$. Assume that $\sum_{n=1}^\infty\mathbb{P}(\frac{|X_n|}{n}\geq\epsilon)=\infty$. The events in the sum are independent, and so by the second Borel-Cantelli lemma we get $\mathbb{P}(\frac{|X_n|}{n}\geq\epsilon\ \ i.o)=1$. But this implies that $\mathbb{P}(\frac{|X_n|}{n}\nrightarrow 0)=1$, a contradiction.