Let $\{X_n\}$ be independent with $P(X_n = n^2) = \frac{1}{n}$ and $P(X_n = -1) = 1 - \frac{1}{n}$.
Show that $\sum_{n=1}^{\infty}X_n = -\infty$ almost surely.
I found that $E[X_n] = n + \frac{1}{n} - 1 \to \infty$. So intuitively it has to tend to $\infty$. but why $-\infty$?
Fix $n$. Let $S_n=X_1+\ldots+X_n$ and let $E_i$ be the event that $X_{i+1},\ldots,X_n$ are all equal to $-1$. Then $Pr[E_i]=\prod_{j=i+1}^{n}\left(1-\dfrac{1}{j} \right)=\dfrac{i}{n}$. Now let $i =\lfloor\sqrt{2n} \rfloor$, and note that if $E_i$ is false, then $S_n > 2n-n=n$. The probability that $S_n>n$ is thus at least the probability that $E_i$ is false, which is $1-\dfrac{\lfloor \sqrt{2n} \rfloor}{n} \to 1$ as $n \to \infty$. Thus, the series diverges to $\infty$ almost surely.