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$X_1, X_2, X_3,...$ independent r.v.'s and $\mathbb{P}(X_n=\frac{-1}{\sqrt{n}})=\mathbb{P}(X_n=0)=\mathbb{P}(X_n=\frac{1}{\sqrt{n}})=\frac13$ check if $$\frac{X_1+X_2+X_3+\dots +X_n}{\sqrt{\log(n)}}$$ converges in distribution, and if it does give the limit distribution.

As a hint I am given: $\lim_{n\to \infty} (1+\frac12+\frac13+\dots+\frac1n-\log(n))=\gamma$ the Euler constant $\gamma=0,577...$

So, I guess I need to use CLT in Lindeberg version. I tried to Consider $Y_n=\frac{X_n}{\sqrt{\log(n)}}$ but then I could not prove Lindeberg condition is satisfied. Maybe a hint how to use a hint would clarify something.

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    $\begingroup$ Maybe a first hint is that Lindenberg's condition is related to the intuitive fact that the variances add to something finite: $\sum_{i=1}^\infty \frac{\mathbb{E}[X_i^2]}{\sqrt{\log (n)}} < \infty$. Can you check the latter condition? Can you see its relationship to Lindenberg's conditon? $\endgroup$ – Kore-N Dec 5 '19 at 21:56

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