Assume we have a sequence of IID r.v.s $X_1, X_2,...,X_n ∼ N(−1, 2^2)$. Calculate the limit in the Probability sense: $$\lim_{n\rightarrow\infty} \frac{n}{X_1^2+X_2^2+...+X_n^2}$$
What I did:
So for first, I inversed the limit and separated it by parts $\rightarrow$ $\frac{X_1^2}{n}+\frac{X_2^2}{n}+...+\frac{X_n^2}{n}$
Then as r.v.s are IID, $E[X_1] = E[X_2] = ... = E[X_n] = -1$ $$E[X_1^2] = E[X_2^2] = ... = E[X_n^2] = \frac{X_1^2}{n} = var(X_1) + (E[X_1])^2 = 4 + 1 = 5$$
$$\frac{X_1^2}{n}+\frac{X_2^2}{n}+...+\frac{X_n^2}{n} = 5 + 5 + ... + 5 = 5n$$ and after inverseing it back we get $$\lim_{n\rightarrow\infty} \frac{n}{X_1^2+X_2^2+...+X_n^2} = \frac{1}{5n} = 0$$
So we get $X=0$, and by definition it is $$P(|X_n - X| \ge \epsilon) = 0$$ $$|X_n| \ge \epsilon$$ but $X_n$ is centered at $X=-1$ so $P(|X_n + 1| \ge \epsilon) = 0$
I am not sure if this is correct, is there anything I have done wrong?
Using SLLN you have that
$$\frac{\sum_iX_i^2}{n}\xrightarrow{a.s.}\mathbb{E}[X^2]=4+1=5$$
thus your sequence converges a.s. to $1/5$ and thus it converges also in probability