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. 
 A: 
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.
