Prove almost sure convergence for log series

Question

If $\{\xi_k, k\geq1\}$ are i.i.d. Show that $$ \frac{\log n}{n} \sum_{k=3}^{n} \frac{\xi_{k}}{\log k} \rightarrow 0 \quad \text { a.s.} $$ if and only if $$ E\left|\xi_{1}\right|<\infty, E \xi_{1}=0 $$

Answer 1

Only partial answer missing one case, but I thought it might be helpful

Let $S_k = \sum_{j=3}^k \xi_j$. Then $$\sum_{k=3}^n \frac{\xi_k}{\ln(k)} = \frac{S_n}{\ln(n)} + S_{n-1}(\frac{1}{\ln(n-1)} - \frac{1}{\ln(n)}) + .... = \frac{S_n}{\ln(n)} + \sum_{k=3}^{n-1} S_k(\frac{1}{\ln(k)} - \frac{1}{\ln(k+1)}) $$

Hence we get $$\frac{\ln(n)}{n} \sum_{k=3}^n \frac{\xi_k}{\ln(k)} = \frac{S_n}{n} + \sum_{k=3}^{n-1} \frac{S_k}{k} \cdot \frac{k\ln(n)\ln(1+\frac{1}{k})}{n\ln(k)\ln(k+1)} $$

Now, assuming $\mathbb E[|\xi_1|]<\infty, \mathbb E[\xi_1]=\mu$, we get $\frac{S_n}{n} \to \mu$ a.e. All that's left is the sum. Given $\varepsilon > 0$ we can find such $N(\varepsilon)$ such that for any $n \ge N(\varepsilon) $ and all $N(\varepsilon) \le k \le n$ we have $\frac{S_k}{k} \in (\mu-\varepsilon,\mu+\varepsilon)$ and $k\ln(1+\frac{1}{k}) \in (1-\varepsilon,1+\varepsilon)$, hence we can bound the sum from below/above by $$\sum_{k=3}^{N(\varepsilon)} \frac{S_k}{k} \cdot \frac{k\ln(n)\ln(1+\frac{1}{k})}{n\ln(k)\ln(k+1)} \pm (1 \pm \varepsilon)(\mu \pm \varepsilon) \frac{\ln(n)}{n}\sum_{k=N(\varepsilon)}^n \frac{1}{\ln(k)\ln(k+1)} $$

In the first sum there are only finitelly many terms and every term goes to $0$ as $n \to \infty$ (with fixed $k \in \{3,...,N(\varepsilon)\}$) while $$ \frac{\ln(n)}{n}\sum_{k=3}^n \frac{1}{\ln(k)\ln(k+1)} \le \frac{\ln(n)}{n}\frac{\sqrt{n}}{\ln(3)\ln(4)} + \frac{\ln(n)}{n} \frac{n-\sqrt{n}}{\ln(\sqrt{n})\ln(\sqrt{n}+1)} \to 0$$

So we get $\frac{\ln(n)}{n} \sum_{k=3}^n \frac{\xi_k}{\ln(k)} \to \mu$ a.e, hence if $\xi_1$ is integrable, we need to have $\mathbb E[\xi_1]=0$ for the limit to be $0$.

Now, if $\xi_1$ is not integrable, we have $3$ cases:

Case 1) $\mathbb E[\xi_1^+]=\infty, \mathbb E[\xi_1^-] = a < \infty$

Then by considering $\xi_k = \xi_k^+ - \xi_k^-$ and linearity of our sum with respect to $\xi_k$ we get $$ \frac{\ln(n)}{n} \Big (\sum_{k=3}^n \frac{\xi_k^+}{\ln(k)} - \sum_{k=3}^n \frac{\xi_k^-}{\ln(k)} \Big )$$ By above, we get that part with $\xi_k^-$ converges to $a$, while part with $\xi_k^+$ will diverge to infinity ($\frac{S_n^+}{n} \to \infty$ and the sum $\sum_{k=3}^n \frac{S_k^+}{k} \frac{k\ln(n)\ln(1+\frac{1}{k})}{n\ln(k)\ln(k+1)}$ is positive ). So our limit is $+\infty$ in this case.

Case 2) $\mathbb E[\xi_1^-]=\infty, \mathbb E[\xi_1^+] = a < \infty$

Really the same reasoning gives us limit $-\infty$ in this case.

Case 3) $\mathbb E[\xi_1^-]=\infty, \mathbb E[\xi_1^+] = \infty$

I don't have a proof yet