The variables $\ X_1, X_2, . . ., X_n \ $ are independent and have a normal distribution with mean $1$ and variance $5$.
$$ Y_n = \frac{|X_1 - X_2| + |X_2 - X_3| + ... + |X_n - X_{n+1}|}{X_1^2 + X_2^2 + X_3^2 +... X_n^2} $$ Is the sequence $Y_n$ convergent almost certainly? If so, what is its limit?
So I thought about using the law of large numbers. We can write that: $$ Y_n = \frac{|X_1 - X_2| + |X_2 - X_3| + ... + |X_n - X_{n+1}|}{X_1^2 + X_2^2 + X_3^2 +... X_n^2} =$$
$$ = \frac{|X_1 - X_2| + |X_2 - X_3| + ... + |X_n - X_{n+1}|}{n} \cdot \left( \frac{X_1^2 + X_2^2 + X_3^2 +... X_n^2}{n} \right)^{-1} $$
We have that: $$\sum_{i =1}^{n} \frac{ Var(X_i) }{n^2} = \sum_{i =1}^{n} \frac{ 5 }{n^2} = 5 \sum_{i =1}^{n} \frac{ 1 }{n^2} < \infty$$
Therefore we can say that by the low of large numbers:
$$\frac{ \sum_{i =1}^{n} X_i^2}{n} \to \mathbb{E}(X^2)$$
And we know that $\mathbb{E}(X^2) = = V(X) + (E(X))^2 = 5 - 1 = 4$
We know that function $ \ f(x) = \frac{1}{x} \ $ is contionus in $4$ so we have $\left( 4 \right)^{-1} = \frac{1}{4}$.
Therefore by the low of large numbers:
$$Y_n = \frac{|X_1 - X_2| + |X_2 - X_3| + ... + |X_n - X_{n+1}|}{n} \cdot \frac{1}{4} $$
But what about the left part? I have no idea what to do next.
Hint: Spilt the sum $|X_1 - X_2| + |X_2 - X_3| + ... + |X_n - X_{n+1}|$ into the sum of even terms and the sum of odd terms. You can apply Strong Law to each of these.
[$|X_1-X_2|, |X_3-X_4|,|X_5-X_6|,...$ is an i.i.d. sequence and so is $|X_2-X_3|, |X_4-X_5|,|X_6-X_7|,...$].