Show almost everywhere convergence for variable with Chi distribution

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Setting

$$X_1,X_2,\ldots \overset{d}{\sim} \mathcal{N}(0,1)$$ $$R_n = \sqrt{X_1^2 + \ldots + X_n^2}$$

I would like to show $\frac{R_n}{\sqrt{n}} \rightarrow 1$ almost everywhere. I have tried to set it up using Borel-Cantelli lemma, where by forming the event $$A_n = \left\{\left|\frac{R_n}{\sqrt{n}} - 1\right| \ge \epsilon \right\}$$

I would show the series $$\sum_{n} \Pr\{A_n\} < \infty$$

and conclude $\Pr\{ A_n, ~i.o.\} = 0$.

So I require $\Pr\{A_n\} \rightarrow 0$ for suitably large $n$. However, I am not sure how to show

$$Pr\left\{\left|\frac{R_n}{\sqrt{n}} - 1\right| \ge \epsilon \right\} \rightarrow 0$$

My problems are

  1. expected value of $\frac{R_n}{\sqrt{n}}$ is 1 in the limiting case only, so what tools can I use here to put a bound on the probability?

  2. I tried taking the fourth moment of $\frac{R_n}{\sqrt{n}} - 1$, but it is not looking good for convergence.

Some directions, both conceptual and relating to algebra would be greatly appreciated!

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If $(X_i)$ are independent, $(X_i^2)$ are also independent. So, by the Strong Law of Large Numbers (check conditions)

$$\frac{1}{n}\sum_{k=1}^n X_k^2\rightarrow^{a.s.} EX_1^2=1$$ Then the continuous mapping theorem implies your result (As pointed by @NateEldredge, continuity is enough here)