$X_i$, $i=1,2,..n$ independent R.Vs $P(X_i=1)=\frac{3}{4} ,\ P(X_i=-1)=\frac{1}{4}$. Prove $\sum_{i=0}^nX_i \to \infty$ a. s. as $n \to \infty$ I am asked to prove $X_1+X_2+X_3+...+X_n$ diverges almost surely as $n \to \infty$ 
Let $Y_n=X_1+X_2+...+X_n$ then what we want to prove is $P(Y_n=k)=1, \text{ as} (k,n) \to (\infty,\infty)$ Let us call $f(n,k)=P(Y_n=k)$
Notice $f(n,k)=f(n-1,k-1)\frac{3}{4}+f(n-1,k+1)\frac{1}{4}$ and $k\leq n$ hence it is sufitient that $f(n,k) \to 1 \text{, as } k \to \infty$ I am stuck here. 
 A: Let $\mu=E[X_{i}]=\frac{1}{2}$. Let $S_{n}=\sum_{i=1}^{n}X_{i}$.
By the strong law of large numbers, there exists a measurable set $\Omega_{0}$
with $P(\Omega_{0})=1$ such that for each $\omega\in\Omega_{0}$,
$\frac{S_{n}(\omega)}{n}\rightarrow\mu$. Now, for any $\omega\in\Omega_{0}$,
\begin{eqnarray*}
 &  & \lim_{n\rightarrow\infty}S_{n}(\omega)\\
 & = & \lim_{n\rightarrow\infty}n\cdot\frac{S_{n}(\omega)}{n}\\
 & = & \infty.\\
\end{eqnarray*}
A: Let $S_n$ denote  $\sum_{k=1}^n X_k$ and let $A>0$. Let us prove that the series $\sum_n P(S_n\leq A)$ is convergent. Note that $S_n=\sum_{k=1}^n 1_{X_k=1}-\sum_{k=1}^n 1_{X_k=-1}=2\sum_{k=1}^n 1_{X_k=1} - n$, hence $\displaystyle \frac{S_n+n}2$ follows $\mathcal B(n,\frac 34)$. Using Hoeffding's inequality, $$P(S_n\leq A)=P(\frac{S_n+n}2\leq \frac{A+n}2)\leq \exp\left(-\frac 2n\left(\frac 34n-\frac{A+n}2 \right)^2\right)=O\left(\exp(-\frac n{8}) \right)$$
Hence $\sum_n P(S_n\leq A)<\infty $ and by Borel-Cantelli lemma, $P(\limsup_n S_n\leq A) = 0$, that is $P(\liminf_n S_n> A)=1$
Note that $P(S_n\to \infty) = P(\bigcap_{A>0} \liminf_n S_n> A)=P(\bigcap_{m\in \mathbb N} \liminf_n S_n> m)$.
But since the countable intersection of almost sure events is almost sure, we already have $P(\bigcap_{m\in \mathbb N} \liminf_n S_n> m)=1$, hence $S_n\to \infty$ a.s.
