Almost sure convergence to mean with a logarithm Let $(X_n)$ be i.i.d. random variables with mean $\mu$. Show that
$$\lim _{n \rightarrow \infty} \frac{1}{\log n} \sum_{i=1}^{n} \frac{X_{i}}{i}=\mu$$
almost surely.
I am thinking of letting $T_n=\frac{1}{\log n} \sum_{i=1}^{n} \frac{X_{i}}{i}$ then applying Markov's inequality to get
$$\Pr(|T_n - \mu|>\epsilon) \leq \frac{\mathbb{E}|T_n-\mu|}{\epsilon^2}$$
and somehow getting a summable sequence to apply the Borel-Cantelli Lemmas. But I don't really know how.

Unfortunately, it seems hard to get a good control on $\mathbb{E}|T_n-\mu|$. 
However, we can use the classical strong law of large numbers. First, using the fact that $\sum_{i=1}^ni^{-1}/\log n\to 1$, we can assume that $\mu=0$. Then denote $S_i=\sum_{k=1}^iX_i$, $S_0=0$  and write 
$$
\frac 1{\log n}\sum_{i=2}^n \frac{S_i-S_{i-1}}i=\frac 1{\log n}\sum_{i=2}^n \left(\frac{S_i}{i}-\frac{S_{i-1}}{i-1}\right)+\frac 1{\log n}\sum_{i=2}^n S_{i-1}\left( \frac{1}{i-1}-\frac 1i\right).
$$
The first sum telescops, for the second one, use the fact that if $c_i\to 0$, then 
$$
\frac 1{\log n}\sum_{i=2}^nc_i/i\to 0.
$$
