Question about the difference between WLLN and SLLN? Let $X, X1,X_2,\cdots$ be i.i.d random.variables. Check if the WLLN and the SLLN hold if the common d.f. is given by,
(1)$P(X=n)=P(X=-n)=\frac{C}{2n^2ln^2(n)}$, $n \geq 3$
(2)$P(X=n)=P(X=-n)=\frac{C}{2n^2ln(n)}$, $n \geq 3$
(3)At this time, $X_i$ are independent though not identical. Let $P(X_1=0)=P(X_2=0)=1$ and $P(X=n)=P(X=-n)=\frac{1}{2nln(n)}$,$P(X_n=0)=1-\frac{1}{nln(n)}$, $n \geq 3$  (Hint:This sequence obeys the WLLN but not the SLLN)
Solution:
According to the SLLN:  Let $X_1,X_2,\cdots$ be pairwise independent with identically distributed random variables with $E|X_i|<\infty$. Let $E(X_1)=u$ and $S_n=X_1+X_2+\cdots+X_n$. Then, $S_n/n \rightarrow u$ as $n\rightarrow \infty$
According to the WLLN:  Let $X_1,X_2,\cdots$ be i.i.d with $E|X_i|<\infty$. Let $E(X_1)=u$ and $S_n=X_1+X_2+\cdots+X_n$. Then, $S_n/n \rightarrow u$ as $n\rightarrow \infty$
As for the question (1):
$E(X_i)=\sum_{k\geq 3}\frac{C}{2k^2ln^2(k)}k+\sum_{k\leq -3}\frac{C}{2k^2ln^2(k)}k=0$
$E|X_i|=\sum_{k\geq 3}\frac{C}{2k^2ln^2(k)}|k|+\sum_{k\leq -3}\frac{C}{2k^2ln^2(k)}|k|=2\sum_{k\geq 3}\frac{C}{2k^2ln^2(k)}|k|=\sum_{k\geq 3}\frac{C}{2k  ln^2(k)}=\infty$
It seems that (1) does not satisfy both WLLN and SLLN. Since $E|X_i|=\infty$\
As for the question (2):
$E(X_i)=\sum_{k\geq 3}\frac{C}{2k^2ln(k)}k+\sum_{k\leq -3}\frac{C}{2k^2ln(k)}k=0$
$E|X_i|=\sum_{k\geq 3}\frac{C}{2k^2ln(k)}|k|+\sum_{k\leq -3}\frac{C}{2k^2ln(k)}|k|=2\sum_{k\geq 3}\frac{C}{2k^2ln(k)}|k|=\sum_{k\geq 3}\frac{C}{2k  ln(k)}=\infty$
It seems that (2) does not satisfy both WLLN and SLLN. Since $E|X_i|=\infty$\
What is the difference between the condition of WLLN and SLLN? Can the statement satisfy the WLLN, though not satisfy SLLN?  Or can the statement satisfies the SLLN, though not satisfy WLLN? For me, the conditions for these two are the same.
Thanks a lot!
 A: This is the standard example of a sequence that satisfies the hypotheses of WLLN but not of SLLN, so you are correct. Also note that WLLN says something about convergence in probability, while SLLN says something about convergence almost surely, which is stronger.
If the hypotheses of SLLN would always be satisfied if those of WLLN are satisfied, WLLN would be a useless result after all. This exercise shows that this needs not be the case.
A: Sequence (1) satisfy WLLN and SLLN since i.i.d.r.v's satisfy $\mathbb E[|X|]<\infty$. 
Consider sequence (2). Since $\mathbb E[|X|]=\infty$, SLLN is not fulfilled. 
This is a consequence of the SLLN:

Let $X_1,X_2,\ldots$ be i.i.d. and $S_n=X_1+\ldots+X_n$. Then $\frac{S_n}{n}\to a$ a.s. iff exists $\mathbb E[X_1]=a$.

And the WLLN can be fulfilled. You can find the following theorem in W.Feller's book "An Introduction to Probability Theory and its Applications", in Vol.2, Chapter VII, paragraph 7: 

Theorem. Let $X_1,X_2,\ldots$ be independent with a common distribution $F$. In order that there exist constants $\mu_n$ such that for each $\epsilon>0$
$$\mathbb P\left(\left|\frac{S_n}{n}-\mu_n\right|>\epsilon\right)\to 0\tag{1}$$
  it is necessary and sufficient that 
  $$n \mathbb P(|X_1|\geqslant n)\to 0 \text{ at } n\to\infty. \tag{2}$$
  In this case (1) holds with 
  $$\mu_n =\mathbb E[X_1\mathbb 1_{\{|X_1|\leq n\}}].$$ 

Since our r.v.'s have symmetric distribution, we can take $\mu_n=\mu=0$ and check only (2). 
$$
n \mathbb P(|X_1|\geqslant n) = n\sum_n^\infty \frac{C}{k^2 \ln k}\leqslant \frac{n}{\ln n}\sum_n^\infty \frac{C}{k^2} \sim \frac{n}{\ln n} \int_n^\infty \frac{C}{x^2}dx = \frac{C}{\ln n}\to 0.
$$
Therefore WLLN fulfilled. 
For (3) both WLLN and SLLN are not fulfilled. You can check it by the same way. Note that the values $X_3,X_4,\ldots $ are i.i.d. so you can apply both theorems. $X_1$ and $X_2$ don't matter. 
