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I have a simple question regarding almost sure convergence. Assume a sequence $X_n = a_1 + ... + a_n$. and $P(a_n \neq -1 \space i.o) = 0$ This means $X_n/n \rightarrow -1 a.s $. Does it also mean $X_n \rightarrow -\infty \space a.s$ ? Thanks.

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  • $\begingroup$ If there exists an $M>0$ such that $|X_n|\leq M$ for all $n$, then $X_n/n$ cannot tend to $1$ almost surely, can it? $\endgroup$ – Stefan Hansen Jan 24 '14 at 20:53
  • $\begingroup$ oh sorry there was a typo i have fixed it. $\endgroup$ – help_needed Jan 24 '14 at 22:14
  • $\begingroup$ My comment was intended to make you come up with a contradiction if you assumed that $X_n$ didn't diverge to $\infty$ (or $-\infty$ now). $\endgroup$ – Stefan Hansen Jan 25 '14 at 7:33
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This follows from the very definition of convergence of sequences: Since

$$\frac{X_n}{n} \to -1 \quad \text{a.s.}$$

we can choose $N=N(\omega)$ such that

$$\left| \frac{X_n(\omega)}{n}- (-1) \right| < \frac{1}{2}$$

for $n \geq N$. In particular,

$$\frac{X_n(\omega)}{n} \leq -\frac{1}{2}$$

Hence

$$X_n(\omega) \leq - \frac{1}{2} n$$

Letting $n \to \infty$ shows $X_n \to - \infty$.

(Alternatively, you can prove it by contradiction, as @StefanHansen suggested. Suppose that there exists some $M>0$ such that $|X_n| \leq M$ for $n$ sufficiently large, then, obviously, $\frac{X_n}{n} \to 0$ and this is a contradiction to your assumption.)

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