Hello,

LLN doesn't suit. Yes intuitively the 1/nē is similar to the 1/n in the LLN. But here we're looking for the convergence in probability, which appears in the small law of large numbers, a barely used theorem.

That's why I think that it isn't the LLN we need here.

After wandering, I'm thinking of using Chebychev's & Jensen's inequalities. The problem is that nowhere will I use the fact that it's a normal distribution... If you added the normal distribution by some mistake, please tell

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Let

Chebychev's inequality can be used because is an independent sequence with a finite second moment.

This gives

Jensen's inequality is usually for convex functions, but for concave functions you just have to reverse the inequality !

So since the square root function is concave, we get that

But we know that the sum the squares of the first n integers when n gets big. So we can say that as n goes to infinity. If you want to prove it properly well you can do the calculations, but I don't think it's necessary.

Hence tends to 0 as n tends to infinity.

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Now let's compute

by independence of the random variables.

For a given k,

(I'll let you write the missing steps, I'm not here to do the details )

So here again we have an equivalent in .

Hence as n goes to infinity.

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Finally, when .

We're almost there. Now we have to prove that Z converges to 0 in probability, by noting that (and having in mind that a.s. hence in probability)

.

- Why ? Because by the triangle inequality, if , then . Thus and hence the above inequality. -

Since and are both positive, .

Hence

which proves that Z converges to 0 in probability.........