Suppose $\{X_k\}_{k\geq 1}$ are independent with
$$P(X_k=k^2)=\frac{1}{k^2}=p_k, P(X_k=-1)=1-p_k.$$
Show $\sum_{k=1}^n X_k\rightarrow -\infty$ almost surely as $n\rightarrow \infty.$
I can see that each $X_k\rightarrow-1$ almost surely, because $P(X_k\rightarrow-1)=1$, and I think that Borel-Cantelli Lemma may be useful but I don't know how to proceed, I would appreciate any hint.
Best Answer
$\sum P(X_k=k^{2})=\sum \frac 1 {k^{2}} <\infty$. Borel cantelli Lemma shows that with probability $1$, $X_k \neq k^{2}$ for all sufficiently large values of $n$. But the only values of $X_k$ are $k^{2}$ and $-1$ so $X_k=-1$ for all sufficiently large values of $n$. This implies that $ \sum\limits_{k=1}^{n} X_k \to -\infty$ with probability $1$.