Let $X_1, X_2, \ldots$ be i.i.d. random variables with finite mean $\mu$.
Prove that
$$
\frac{1}{\ln n} \sum_{k=1}^n \frac{X_k}{k} \to \mu
\text{ a.s.}
$$
I was given the hint to look at the subsequence $2^{2^n}$, but I'm not seeing how that helps. We can almost apply the strong law of large numbers here, but not quite.
Best Answer
Let $S_r = \sum_{k=1}^r X_k$ and for easier notation, put $S_0 = 0$
Note that $$ \sum_{k=1}^n \frac{X_k}{k} = \sum_{k=1}^n \frac{S_k-S_{k-1}}{k} = \sum_{k=1}^n \frac{S_k}{k} - \sum_{k=1}^n \frac{S_{k-1}}{k} = \frac{S_n}{n} + \sum_{k=1}^{n-1} \frac{S_k}{k} - \sum_{k=2}^n \frac{S_{k-1}}{k} = \frac{S_n}{n} + \sum_{k=1}^{n-1} \frac{S_k}{k} - \sum_{k=1}^{n-1} \frac{S_k}{k+1} = \frac{S_n}{n} + \sum_{k=1}^{n-1}S_k(\frac{1}{k} - \frac{1}{k+1}) = \frac{S_n}{n} + \sum_{k=1}^{n-1} \frac{S_k}{k}\frac{1}{k+1}$$
Put $W_r = \frac{S_r}{r}$, so that we have $$\frac{1}{\ln(n)} \sum_{k=1}^n \frac{X_k}{k} = \frac{W_n}{\ln(n)} + \frac{1}{\ln(n)} \cdot \sum_{k=1}^{n-1} \frac{W_k}{k+1}$$
Call $Y_n = \frac{W_n}{\ln(n)}$. Since by SLLN $W_n$ tends to $\mu$ a.e, so $Y_n$ tends to $0$ a.e
Now, as we said, $W_n$ tends to $\mu$ a.s. So we have set $\Omega_0$, $\mathbb P(\Omega_0) =1$, such that for $\omega \in \Omega_0$ $W_n(\omega) \to \mu$
Take that $\omega \in \Omega_0$. It is sufficient to show that $\frac{1}{\ln(n)}\sum_{k=1}^{n-1} \frac{W_k(\omega)}{k+1}$ tends to $\mu$.
We need $1$ common fact from analysis, that for any $N \in \mathbb N$:
$$ \lim_{n \to \infty} \frac{\sum_{k=N}^n \frac{1}{k}}{\ln(n)} = 1$$
Now, take any $\epsilon>0$ take such $N \in \mathbb N$ that for $n>N$ we have $|W_n(\omega) - \mu| <\epsilon$.
Then $$ \frac{1}{\ln(n)} \sum_{k=1}^n \frac{W_k(\omega)}{k+1} = \frac{1}{\ln(n)}\sum_{k=1}^N \frac{W_k(\omega)}{k+1} + \frac{1}{\ln(n)} \sum_{k=N}^n \frac{W_k(\omega)}{k+1}$$
The first one has only finitelly many terms in the sum, so tends to $0$ as $n \to \infty$ (note $N$ is fixed)
For the second one, we can bound it from below and above:
$$ \frac{\mu - \epsilon}{\ln(n)} \sum_{k=N}^n \frac{1}{k+1} \le \frac{1}{\ln(n)}\sum_{k=N}^n \frac{W_k(\omega)}{k+1} \le \frac{\mu + \epsilon}{\ln(n)}\sum_{k=N}^n \frac{1}{k+1}$$
Now use our fact from analysis, to conclude that for our $\omega \in \Omega_0$, we have:
$$ \mu -\epsilon \le \liminf \frac{1}{\ln(n)} \sum_{k=N}^n \frac{W_k(\omega)}{k+1} \le \limsup \frac{1}{\ln(n)} \sum_{k=N}^n \frac{W_k(\omega)}{k+1} \le \mu+\epsilon$$
Since $\epsilon >0$ was arbitrary, we have $$\lim \frac{1}{\ln(n)} \sum_{k=1}^n \frac{W_k(\omega)}{k+1} = \mu$$Again, since $\omega \in \Omega_0$ was arbitrary, we have $$\lim \frac{1}{\ln(n)} \sum_{k=1}^n \frac{W_k}{k+1} = \mu \ \ a.e $$
Using everything we proved above, we can conclude: $$ \lim_{n \to \infty} \frac{1}{\ln(n)} \sum_{k=1}^n \frac{X_k}{k} = \mu \ \ a.e $$