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?

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.

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?

Thanks.

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?