# Almost sure convergence of independent Gaussians implies convergence of means and variances

Let $$\{X_n\}_{n \geq 1}$$ be a sequence of independent random variables with $$X_n \sim N(\mu_n, \sigma_n^2)$$ for each $$n \geq 1$$. Show that $$S_n = \sum_{i = 1}^n X_i$$ converges almost surely if and only if both $$\sum_{n = 1}^\infty \mu_n$$ and $$\sum_{n = 1}^\infty \sigma_n^2$$ converge.

This has been answered here. The direction $$\sum_{n = 1}^\infty \mu_n < \infty$$ and $$\sum_{n = 1}^\infty \sigma^2 < \infty$$ implying $$S_n$$ converges almost surely is clear. I would like to prove the other direction without using characteristic functions, as this problem has been assigned to me in a course where we haven't encountered them yet.

Almost sure convergence implies convergence in distribution, so it suffices to show the $$S_n$$ do not converge in distribution. We then have
$$S_n \sim N\left(\sum_{i=1}^n \mu_i, \sum_{i=1}^n \sigma^2_i\right),$$
• The $S_n$ converge in distribution iff their distribution functions $F_n \to F$ pointwise, which obviously occurs iff the sums converge. @SayanDutta