Divergence in probability implies almost sure divergence

Question

Let $X_n$, $Y_n$ be dependent random variables

with $X_n \to c$ almost surely with $c>0$ and $\log(n) Y_n \to 0$ in probability.

Can I state the following:

$\log(n) (X_n + Y_n) \to \infty$ almost surely, i.e. the term diverges almost surely?

Answer 1

No. Let $\{Y_n\}$ be independent random variables such that $P(Y_n=-1)=\frac 1 n$ and $P(Y_n=0)=1-\frac 1 n$. Let $X_n=1$ for all $n$. Note that $P(|\log(n)Y_n|>\epsilon)\leq \frac 1 n \to 0$ for any $\epsilon >0$. Since $\sum P(Y_n=-1)=\sum \frac 1 n=\infty$ it follows by Borel Cantelli Lemma that $Y_n=-1$ infinitely often with probability $1$. Hence $log(n)(X_n+Y_n)=0$ infinitely often with probability $1$.