Solution
= Solution
{c}
There are exactly $n$ summands in $S_n$, and the <triangle inequality> gives
$$
\mathbb E\left|\frac{S_n}{n}-1\right|
\leq\frac1n\sum_{i=2}^{n+1}\mathbb E|X_i-1|
=\frac1n\sum_{i=2}^{n+1}\frac1{\log i}\longrightarrow0.
$$
The last step is convergence of arithmetic means of a sequence tending to zero. Thus $S_n/n\to1$ in $L^1$, and <Markov's inequality> gives
$$
\boxed{S_n/n\longrightarrow1\text{ in probability}.}
$$