Let $(X_n)_n$ be a sequence of i.i.d random variables.
Prove that the following statements are equivalent:
a) $X_1 \in L^1.$
b) $\forall x \in \mathbb{R},\frac{1}{n}\sum_{k=1}^n\sin(kx)X_k$ converges a.s.
c) $\exists x \in \mathbb{R}-\pi\mathbb{Z}; \frac{1}{n}\sum_{k=1}^n \sin(kx)X_k$ converges a.s
a) Prove that $X_1 \in L^2$ if and only if $\exists x \in \mathbb{R}-\pi\mathbb{Z} $ ans a sequence $(x_n)_n$ of real numbers such that $\frac{1}{\sqrt{n}}\sum_{k=1}^n\sin(kx)X_k-x_n$ converges in distribution.
Hint: Use characteristic functions. You may consider the random variables $Y_n=\frac{1}{\sqrt{n}}\sum_{k=1}^n\sin(kx)(X_{2k+1}-X_{2k}),$ for the converse.
b) Deduce that $\forall p>2,x \in \mathbb{R},\frac{1}{n^{1/p}}\sum_{k=1}^n\sin(kx)X_k$ diverges a.s.
For the first question, suppose that $X_1 \in L^1,$ I used Kolmogorov two series theorem, to prove that $\sum_{n}\frac{1}{n}\sin(nx)(X_{n}-E[X_1])1_{|X_n-E[X_1]| \leq n}$ converges a.s, so let $Y_n=\frac{1}{n}\sin(nx)(X_{k}-E[X_1])1_{|X_n-E[X_1]| \leq n},$
$\sum_nP(Y_n \neq \frac{1}{n}\sin(nx)(X_n-E[X_1]))=\sum_{n}P(|X_1-E[X_1]| \geq n)<+\infty,$
$\sum_n Var(Y_n-E[Y_n]) \leq \sum_n \frac{1}{n^2}E[(X_1-E[X_1])^2 1_{|X_1-E[X_1]| \leq n}]<+\infty,$ since $E[|X_1-E[X_1]|]<+\infty$ which implies that $\sum_{n}(Y_n-E[Y_n])$ converges a.s, further $\sum_{n}|E[Y_n]| \leq \sum_n P(n<|X_1-E[X_1]| \leq n+1)\sum_{k=1}^n\frac{|\sin(kx)|}{k} \leq x\sum_{n}nP(n<|X_1-E[X_1]| \leq n+1)<+\infty$
so $\sum_n\frac{\sin(nx)}{n}(X_n-E[X_1])$ converges a.s and since $\sum_n \frac{1}{n}\sin(xn)$ converges, we conclude that $\sum_{n}\frac{1}{n}\sin(nx)X_n,$ converges a.s, then using Kronecker lemma $\frac{1}{n}\sum_{n}\sin(nx)X_n$ converges a.s to $0$.
To prove that $X_1 \in L^1,$ I took a subsequence $n_k=\left \lfloor{\frac{5\pi}{6}+2\pi k}\right \rfloor$ such that $\frac{1}{2} \leq \sin(n_kx)$ and then we will have $\frac{1}{n}X_{n_k}$ converges a.s to $0$, which means that $X_1 \in L^1.$
Needing help for 2) a), part b) is a conclusion from a).
Appreciating any ideas.