Almost sure convergence of sum of independent Bernoullis Let $X_n$ be independent random variables where $X_n \sim Bernoulli(\frac{1}{n})$. I'd like to show that $Y_n = \frac{1}{\log(n)}\sum_n{X_n}$ converges to $1$ almost surely.
My idea is to first show that $Y_n$ converges to $1$ in probability, and then to use Kolmogorov's convergence criterion to show that the convergence is almost sure as well.
However, I'm running in problems even in showing the convergence in probability. I'm trying to use the bound $\log(n) + \frac{1}{n} \leq \sum_{j=1}^{n}{\frac{1}{j}} \leq \log(n) + 1$ in tandem with Markov's inequality, but the absolute value in the Markov inequality is killing me.
$P(|Y_n - 1| \geq \epsilon) \leq \frac{\mathbb E[|Y_n - 1|]}{\epsilon} \leq \frac{\mathbb E[Y_n] + 1}{\epsilon}$, but of course this bound is not good enough. If I didn't have to worry about the absolute values, I would be able to use the upper bound from above, but I'm not sure how to proceed at this point.
Does this look like a logical approach? I think that the second half of the proof might be a bit easier, since Kolmogorov's criterion just needs me to check that the sums of the means and variances converge...
 A: Too long for a comment, only partial answer:
You had a good idea about using inequalities, because in fact convergence almost surely is really connected with probabilities $$ \mathbb P(|Y_n - 1| > \varepsilon)$$ The key point is to use proper inequality here. You said Markov, which gives you bound of type $$ \frac{\mathbb E[|Y_n-1|]}{\varepsilon} $$ which is rather hard to deal with due to this absolute value. It can be shown that $$ \mathbb E[Y_n] = \frac{1}{\ln(n)} \sum_{k=1}^n \mathbb E[X_k] = \frac{1}{\ln(n)} H_n $$ where $H_n$ is $n'$th Harmonic number. Having said that, we would rather compare $$ \mathbb P(|Y_n -1 | > \varepsilon) \le \mathbb P(|Y_n - \frac{H_n}{\ln(n)}| > \eta)$$ for apriopriate $\eta > 0$. But which one would work? Let's take a look. If $|Y_n - \frac{H_n}{\ln(n)}| > \eta$, then $ |Y_n - 1| + |1-\frac{H_n}{\ln(n)}| > \eta$, too, due to triangle inequality. So that $\eta = \varepsilon - |1 - \frac{H_n}{\ln(n)}| $ will be a good choice (we can always (for fixed $\varepsilon >0$) find $N$ big enough such that for $n>N$ our $\eta$ is bigger than $0$ and everything's okay.
This is crucial, because from $\mathbb P(|X-a|>t)$ for some $a$, we arrived at $\mathbb P(|X-\mathbb E[X]| > s)$ which gives us a big opportunity to use Chebyshev.
Hence $$ \mathbb P(|Y_n - 1| > \varepsilon) \le \mathbb P(|Y_n - \mathbb E[Y_n]| > \eta_n) \le \frac{Var(Y_n)}{\eta_n^2} = \frac{Var(\sum_{k=1}^n X_k)}{\ln^2(n)\eta_n^2} = \frac{H_n}{\ln^2(n)\eta_n^2}$$
which gives us convergence in probability.
Not connected to convergence almost surely, but Note that due to Lyapunov inequality ($\mathbb E[|Z|]^2 \le \mathbb E[Z^2]$) we get$$ \mathbb E|Y_n-1| \le \mathbb E|Y_n - \frac{H_n}{\ln(n)}| + |\frac{H_n}{\ln(n)}-1| \le \sqrt{Var(Y_n)} + |\frac{H_n}{\ln(n)}-1| \to 0$$ so $Y_n$ converges to $1$ in $L_1$.
