Does $Z_n=\sum_{k=1}^{n}\sqrt{k}X_k$ satisfy the strong law of large numbers if $ X_n: \begin{matrix}-\frac{1}{n} & \frac{1}{n} \\ \frac{1}{2} & \frac{1}{2} \end{matrix}, n=1,2,...$ are independent random variables.
I have the following theorems, but I cannot prove this, I have tried all which I understand, the theorems I have are:
1.) Strong Law of large numbers states that the sequence $X_1,X_2,...$ must satisfy:
$$\frac{1}{n}\sum_{k=1}^{n}(X_k-EX_k)\to^{a.s.}0, n\to \infty$$
2.) Kolmogorov Law: If $(X_n)$ independent random variables, such that $\sum_{n=1}^{\infty} \frac{\text{Var}(X_n)}{n^2}<\infty$, then the strong law of large numbers is satisfied.
3.)Borels: If $ S_n:\mathcal B(n,p)$ (binomial distribution), then $$\frac{S_n}{n}\to^{a.s.}p, n \to \infty$$
or the consequence:
Let $X_n$, sequence of independent random variables, equally distributed, such that $EX_k=a$ and $\text{Var}X_k= \omega^2, k=1,2,3... \implies$
$$\frac{1}{n}\sum_{k=1}^{n}X_k\to^{a.s.}a, n\to \infty$$
Consider the random variables $Y_n=\sqrt{n}X_n$. Then note $Y_n$ are independent, and $E(Y_n)=0$. Further note that $\sum_{n=1}^\infty Var(Y_n)=\infty$.
Now invoke Kolmogorov's Three Series Theorem. One of the converging series must be $\sum_nVar(Y_n^c)$ for every $c>0$ if $\sum_nY_n<\infty$ almost surely.
Here, $Y_n^c=Y_n$ if $|Y_n|\leq c$ and $0$ otherwise.
Note that $Y_n=\pm\dfrac{1}{\sqrt{n}}$ each with prob. $1/2$ so for any given $c$, there exists $N\in\mathbb N$ such that $|Y_n|\leq c$ for all $n\geq N$.
Hence for sufficiently large $n$ we must have $Y_n^c=Y_n$.
So $\sum_nVar(Y_n^c)\geq \sum_{n\geq N}Var(Y_n^c)=\sum_{n\geq N}Var(Y_n)=\infty$.
Hence almost sure convergence of $\sum_{n}\sqrt{n}X_n$ does not happen.