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We know that:

  1. $\sum \limits _{n=1}^{\infty}\frac{1}{n} = \infty$
  2. $\sum \limits _{n=1}^{\infty}\frac{(-1)^n}{n} < \infty$

Both are easy to show. In the first case we can use the criterion based on integrating the function $f(x) = \frac{1}{x}$ and the convergence of the second series can be settle with the Dirichlet's criterion.
Let's imagine a random series: $$\sum \limits _{n=1}^{\infty}\frac{\lambda_n}{n},$$ where $\lambda = 1$ or $\lambda = -1$.
$P(\lambda = 1) = P(\lambda = -1) = \frac{1}{2}$.
When the series will converge?

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  • $\begingroup$ I guess that it converges...but I don't know why ;) $\endgroup$
    – user
    May 8, 2018 at 22:30
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    $\begingroup$ This section on wikipedia might be interesting and relevant. $\endgroup$
    – B. Mehta
    May 8, 2018 at 22:40

1 Answer 1

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If the random variables $\lambda_n$ are independent and take the values $1$ and $-1$ with probability $1/2$ each then $\sum \lambda_na_n$ converges almost surely if and only if $\sum a_n^{2}$ is finite. In particular this holds for $a_n=1/n$. However the series may not converge almost surely if independence is dropped: if all the $\lambda_n$ are equal then the series diverges almost surely.

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    $\begingroup$ Kolomogorov's three series theorem gives a proof of the claim I have made in my answer. $\endgroup$ May 9, 2018 at 5:46
  • $\begingroup$ @Did Thanks for editing and removing the obvious error I had made. $\endgroup$ May 10, 2018 at 5:06
  • $\begingroup$ Thank you! I will try to prove your claim using Kolomogorov's theorem $\endgroup$
    – Hendrra
    May 11, 2018 at 20:21

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