Determine whether sequence of independent rvs satisfy SLLN 
Determine if the following sequences of independent rvs satisfy the SLLN:
  
  
*
  
*$(a)$ $P(X_n=1) = \frac{1}{n^2},\, P(X_n=0) = 1-\frac{1}{n^2}$
  
*$(b)$ $P(X_n=n) = \frac{1}{n^2},\, P(X_n=0) = 1-\frac{1}{n^2}$
  
*$(c)$ $P(X_n=n) = \frac{1}{n}, \, P(X_n=0) = 1-\frac{1}{n}$
  

My attempt: Let $S_n = X_1 + X_2+\ldots + X_n$.


*

*$(a)$: $E(X_n) = \frac{1}{n^2} = E(X_n^2)$.  We need to verify: $\displaystyle \lim_{n\rightarrow \infty} \frac{S_n - E(S_n)}{n}\rightarrow 0$ almost surely. Now, $E(S_n) = \sum_{k=1}^{n} \frac{1}{k^2}$, so as $n\rightarrow \infty$, $E(S_n)\rightarrow \frac{\pi^2}{6}$, which implies $\frac{E(S_n)}{n}\rightarrow 0$. However, $S_n\leq n$ for every $n$, thus as $n\rightarrow \infty$, $\frac{S_n}{n} = \frac{n-k}{n} \rightarrow 1$, unless all $X_i$'s $=0$ (which is an extreme case, since $X_n$ are random). Thus $X_n$ does not satisfy SLLN. 

*$(b)$: $E(X_n) = \frac{1}{n}$ and $E(X_n^2) = 1$. Thus $E(S_n)$ diverges as $n\rightarrow \infty$, while $S_n = kn$ for some nonnegative integers $k$. Thus, we could apply the L'Hospital rule (variable is $n$) to compute $\lim_{n\rightarrow \infty} \frac{S_n - E(S_n)}{n} = \lim_{n\rightarrow \infty} (\sum_{k=1}^{n} \frac{1}{k^2}) = \frac{\pi^2}{6}\neq 0$, so $X_n$ does not satisfy SLLN.

*$(c)$: $E(X_n) = 1$, so $E(S_n)=  n$. Thus, $S_n - E(S_n) = (k-1)n$ for some nonnegative integers $k$. Thus, $\lim_{n\rightarrow \infty} \frac{S_n - E(S_n)}{n} = k-1 = 0$ if and only if $k=1$. So, $\lim_{n\rightarrow \infty} \frac{S_n - E(S_n)}{n}$ does not converge to $0$ almost surely (is this correct?). Thus, $X_n$ does not satisfy SLLN.
My question: I'm quite skeptical with my solution above since the way I tried to express $S_n$ seems to be weird. Could anyone please help me with these problems in case my solutions above are incorrect? Any thoughts would really be appreciated.
 A: The SLLN by Kolmogorov for independent sequence $\left(X_n\right)_{n\in \mathbb N}$, states that if


*

*$Var(X_n)<\infty, \mathbb E[X_n]=μ_n$ for every $n$,

*$\sum_{n=1}^\infty\dfrac{Var(X_n)}{n^2}<\infty$ 


then $\frac1n(S_n-\mathbb ES_n)\overset{a.s.}\longrightarrow 0$. These are sufficient but not necessary meaning, that every sequence that satisfies these (actually the second is the one to check) satisfies the SLLN, but if a sequence does not satisfy these, then this does not mean that it cannot satisfy the SLLN.
So, in your case:


*

*$(a)$: $\mathrm{Var}[X_n]\le \mathbb E[X_n^2]=\frac1{n^2}$ and so $$\sum_{n=1}^{\infty}\frac{Var(X_n)}{n^2}\le\sum_{n=1}^\infty\frac1{n^4}<\infty$$
and so, it does satisfy the SLLN.

*$(b)$: $\mathrm{Var}[X_n]\le \mathbb E[X_n^2]=1$ and so, as in $(a)$ 
$$\sum_{n=1}^{\infty}\frac{Var(X_n)}{n^2}\le\sum_{n=1}^\infty\frac1{n^2}<\infty$$
and so, it does satisfy the SLLN.

*$(c)$: Here $Var(X_n)=n-1$, hence $$\sum_{n=1}^{\infty}\frac{Var(X_n)}{n^2}=\sum_{n=1}^\infty\frac{n-1}{n^2}=\infty$$
and so, this way does not work! (But as mentioned above, it does not mean that we are done). However, if we consider the independent events $E_n:\{ω:X_n(ω)=n\}$, we have that $$\sum_{n=1}^\infty P(E_n)=\sum_{n=1}^\infty\frac1n=\infty$$ and hence by the second Borel-Cantelli lemma: $$P(\lim\sup_nE_n)=P(X_n=n \text{ occurs i.o.})=1$$ This, however, still does not allow you to show that \begin{align}P\left(\lim_{n}\frac{S_n-\mathbb E[S_n]}{n}\neq 0\right)&=P\left(\lim_{n}\frac{S_n}{n}\neq 1\right)\\[0.2cm]&\ge P\left(\lim_{n}\frac{S_n}{n}> 1\right)= P\left(\lim\sup E_n\right)=1\end{align}
since there are unbounded sequences that are cesaro-summable. 

