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I have a question about conclusion of a problem. For a sequence $(X_k)_{k\in \mathbb{N}}$ of independent random variables with uniform distribution on $[-1,1]$, for all $k\in \mathbb{N}$ and $Y_n = \frac{1}{n}\sum_\limits{k=1}^n \sqrt{k} X_k$, show that that $Y_n \to 0$ almost surrely.

I tried to use the Kolmogorov's Theorem for the conclusion of Strong law of large numbers, but the series $\sum_\limits{k=1}^\infty \frac{D^2(\sqrt{k}X_k)}{k^2} = \sum_\limits{k=1}^\infty \frac{kD^2(X_k)}{k^2} = \sum_\limits{k=1}^\infty \frac{1}{3k}$ diverges. So, I can't use this Theorem. Are there other similar results with Kolmogorov's Theorem? or maybe the conclusion of the problem is wrong.

At the next point of the problem I must to compute: $\lim_\limits{n\to\infty} [\frac{n^n}{\sqrt{n!}}\sin{\frac{\sqrt{1}}{n}}\sin{\frac{\sqrt{2}}{n}}\dots\sin{\frac{\sqrt{n}}{n}}]$. I used characteristic function to solve it, but I want to know if I need a.s. convergence, or is enough convergence in probability(i.e. WLLN) to pass to convergence in characteristic function.

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    That theorem will only prove SLLN in the finite variance case. You need something stronger https://en.wikipedia.org/wiki/Kolmogorov's_three-series_theorem – 2017-01-25
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    Kolmogorov three series theorem shows that the sequence does not converge in the almost sure sense. Were you asked to show its convergence? – 2017-01-26

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The conclusion that you would prove is not true. The reason is the following. Let $X_{nk}=\frac{\sqrt{k}X_k}{n}$, $1\le k\le n$, then $\mathbb{E}X_{nk}=0$, \begin{gather} \sum_{k=1}^n\mathbb{D}(X_{nk})=\sum_{k=1}^n\frac{k}{3n^2}\to \frac16,\qquad \text{as $n\to\infty$, }\\ \max_{1\le k\le n}|X_{nk}|\le \frac1{\sqrt{n}}\to 0.\qquad\text{as $n\to\infty$, } \end{gather} Hence using Lévy-Lindeberg Theorem, $$ \text{d-}\lim_{n\to\infty}Y_n=\text{d-}\lim_{n\to\infty}\sum_{k=1}^nX_{nk} \overset{dist.}{=}N(0,1/6).$$ This also means that ''$Y_n\to 0$ a.s.'' is not true. But you could prove that $$ \frac{Y_n}{\log\log n}\to 0,\qquad \text{a.s.}$$

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    Thank you for your answer! I understand now that the conclusion of the problem was wrong. – 2017-01-27