Convergence of sum of normal random variables with variance $\frac{1}{\sqrt{i}}$

Question

I'm trying to using Kolmogorov's 3 series to show that if

$$ X_i \quad iid \sim N(0, \frac{1}{\sqrt{i}} ) $$

Does

$$ \sum_i^\infty X_i \quad \text{converge?} $$

Given that the sum of the variances $\sum_1^\infty \frac{1}{i^\frac{1}{2}}$ diverges, I'm thinking no it doesn't converge.

Is there a way to show this using Kolmogorov's 3 series?

As per the comment, I tried the following.

$$ P(|X_i| > A) = 2 \int_A^\infty i^{\frac{1}{4}} \frac{1}{\sqrt{2\pi}} e^{-\frac{i^{\frac{1}{2}}}{2} x^2} dx $$

take $y = i^{1/4}x$

$$ P(|X_i|>A)= 2 \int_{Ai^{1/4}}^\infty \frac{1}{\sqrt{2\pi}} e^{-\frac{1}{2} y^2} dy $$

Which is a decreasing function of i.

Answer 1

First of all, since the distribution of $X_i$ depends on $i$, the sequence of variables is not IID. Rather, you have independent but not identically distributed random variables. In any case, to look at convergence, let's examine the partial sums:

$$S_n \equiv \sum_{i=1}^n X_i.$$

Since the underlying variables are independent normal random variables, we have:

$$S_n \sim \text{N}(0, V_n) \quad \quad \quad \quad \quad V_n \equiv \sum_{i=1}^n i^{1/2}.$$

Consequently, for any $s \geqslant 0$ we have:

$$\begin{align} \mathbb{P}(|S_n| > s) &= \mathbb{P} \bigg( \frac{|S_n|}{\sqrt{V_n}} > \frac{s}{\sqrt{V_n}} \bigg) = 2 \Phi \bigg( - \frac{s}{\sqrt{V_n}} \bigg), \\[6pt] \end{align}$$

and since $\lim_{n \rightarrow \infty} V_n = \infty$ we then have:

$$\begin{align} \lim_{n \rightarrow \infty} \mathbb{P}(|S_n| > s) &= 2 \Phi (0) = 2 \cdot \frac{1}{2} = 1. \\[6pt] \end{align}$$

Thus, we can see that for any finite value $s \geqslant 0$ the probability that $|S_n| > s$ will converge to one as $n \rightarrow \infty$. In this sense, the limiting sum of the underlying random variables "explodes" (i.e., it does not converge).