Let $Y_k$ be independently Bernoulli distributed rvs holding $P(Y_k = 1) = \frac{1}{k}$ for every $k = 1,2,...,n$. I need to prove that the sum $\frac{1}{\log n} \sum_{k=1}^n Y_k$ converges in probability to $1$. How can I also check if it converges or not almost surely?
Well, I tried to use some well known inequalities but they don't work. Also Kolmogorov's theorem is of no use here.
As you've mentioned in your comment convergence in probability is not interesting anymore. So, first we show that the series is indeed convergent in $\mathbf{R}$ a.s.
Denote $\tilde{Y}_k := \frac{Y_k}{\log n}$. To use Kolmogorov's two-series theorem we need to show that the series of expected values and variances converge in $\mathbf{R}$.
For the mean we have $$ \sum_{k=1}^n \mathbb{E}\tilde{Y}_k = \frac 1 {\log n} \left(1 + \frac 1 2 + \dots + \frac 1 n\right) \le C_n, $$ with $C_n \to 1$ as $n \to \infty$.
To see the latter inequality, note that $1 + \frac 1 2 + \dots + \frac 1 n$ is a harmonic series, which is known to have a pretty good approximation of $\log [n] + \gamma$, where $\gamma$ is Euler's constant. This approximation gets arbitrary small when $n$ gets larger. Hence, we have $$ \frac 1 {\log n} \left(1 + \frac 1 2 + \dots + \frac 1 n\right) \approx \frac{\log n + \gamma}{\log n} = 1 + \frac{\gamma}{\log n}, $$ so taking $C_n = c\left( 1 + \frac{\gamma}{\log n}\right)$ with some absolute constant $c$ large enough ensures the inequality for the sum of expected value. Hence, $\sum_{k=1}^{\infty} \mathbb{E}\tilde{Y}_k$ converges in $\mathbf{R}$.
Similarly, $$ \sum_{k=1}^n \mathbb{V}ar\tilde{Y}_k = \frac 1 {\log^2 n} \left(\underbrace{1 + \frac 1 2 + \dots + \frac 1 n}_{\log [n] + \gamma} - \underbrace{\left(1 + \frac 1 {2^2} + \dots + \frac 1 {n^2}\right)}_{\pi^2/6}\right) \to 0, \quad \text { as } n \to \infty. $$
So, the theorem implies that $\sum_{k=1}^{n} \frac 1 {\log n} Y_k$ converges a.s. when $n \to \infty$. Finally, since we know that convergence almost surely implies convergence in probability and there is convergence in probability to a constant $1$, then we have almost surely convergence to $1$ as well.