Weak LLN vs Strong LLN

Heres an example of a sequence of r.vs $$(X_n)$$ that obey the Weak LLN but do not obey the Strong LLN, $$P(X_n=\pm n)=1/(2n\ln n)$$ for any $$n=1,2\dots$$ and $$P(X_n=0)=1-1/(n\ln n)$$.

I am trying to show why this is so, but I am having trouble with it. We don't have any closed form expressions of the partial sums of $$(X_n)$$, only the probabilities that each $$X_n$$ takes at $$n$$. How do you even compute $$\sum_n^NX_n/N$$?

With regards to the strong Law, it suffices to show that $$E[X_1]$$ is infinite, but we dont even have that they are identically distributed?

Any help would be greatly appreciated.

• So, what definition of the SLLN do you use? – Will M. Dec 1 '18 at 18:30

Assuming OP means to ask whether or not the random series $$Z_N = \sum\limits_{n = 2}^N \dfrac{X_n}{N}$$ converges almost surely. And that OP assumes $$(X_n)_{n \geq 2}$$ is an independent sequence (we also notice $$X_1$$ makes no sense).
Abridged proof: we fundamentally apply Lindeberg-Feller Central Limit Theorem for triangular array with row-independent random variables (see 8.3 here https://www.ssc.wisc.edu/~xshi/econ715/Lecture_9_local_power.pdf). Our triangular array is $$U_{N, k} = \dfrac{X_k}{N}.$$ Observe $$\mathbf{E}(U_{N, k}) = 0$$ and $$\mathbf{V}\mathrm{ar}(U_{N, k}) = \dfrac{k}{N^2 \log k}.$$ Hence, for $$Z_N = \sum\limits_{k = 2}^N U_{N, k}$$ we have $$\sigma_N^2 = \mathbf{V}\mathrm{ar}(Z_N) = \sum\limits_{k = 2}^N \dfrac{k}{N^2 \log k}$$ and this series clearly converges to some positive number $$\sigma^2.$$ Finally, the LFCLT holds because $$\sum_{k = 2}^N \mathbf{E} \left(U_{N, k}^2 \mathbf{1}_{\left\{|U_{n, k}| > \frac{\varepsilon}{2} \sigma \right\}} \right) \leq \sum_{\frac{\varepsilon}{2} \sigma N \leq k \leq N} \dfrac{k}{N^2 \log k} \leq \dfrac{N(1 - \frac{\varepsilon}{2} \sigma) \times N}{N^2 \log (\frac{\varepsilon}{2} \sigma N)} \to 0.$$ Therefore, $$Z_N \to \mathrm{Norm}(0; \sigma^2).$$ Q.E.D.
Ammend. What the Lindeberg-Fellet CLT theorem gives is weak convergence. However, we know that convergence almost surely implies weak convergence; in other words, if the random variables $$Z_N$$ converges almost surely to any limit (which I did not prove), then said limit is normally distributed. Unfortunately,
$$Z_N$$ will not, in fact, converge almost surely to any limit.
My reasoning follows from the fact that $$Z_{N + 1} = \dfrac{N}{N+1}Z_N + \dfrac{X_{N + 1}}{N + 1}$$ so that, assuming $$Z_N \to Z$$ almost surely, for some (finite-valued) limit $$Z,$$ then $$\dfrac{X_N}{N} \to 0$$ almost surely. However, the random variable $$\dfrac{X_N}{N}$$ only has three values: $$-1,$$ $$0$$ and $$1;$$ whence, if the sequence $$\dfrac{X_N}{N}$$ converges to zero, it has to be zero eventually. In other words, what I argumented was that $$\left\{ \lim_{N \to \infty} \dfrac{X_N}{N} = 0 \right\} \subset \bigcup_{N = 1}^\infty \bigcap_{n = N}^\infty \{X_n = 0\} \mathop{=}^{\mathrm{def}} \bigcup_{N = 1}^\infty \mathrm{J}_N.$$ By independence, $$\mathrm{J}_N$$ is a null event, hence, except on a null set, no point satisfy $$\dfrac{X_N}{N} \to 0$$ and therefore, $$Z_N$$ does not converge on any subset of positive probability. Q.E.D.