# Strong Law of Large Numbers for Harmonic-Like Series

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.

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: $$\lim_{n \to \infty} \frac{1}{\ln(n)} \sum_{k=1}^n \frac{X_k}{k} = \mu \ \ a.e$$