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