If $\lim_{n \rightarrow \infty} \frac{S_n^4}{n^4} = 0$ then $\lim_{n \rightarrow \infty} \frac{S_n}{n} = 0$ where $S_n$ is the sum of $n$ iid RVs with mean zero.

My question

I'm having trouble understanding the final part of a proof for the strong LLN. See red box in screenshot below. I'm thinking the explanation is very quick and obvious but ugh I forgot most of HS Calculus. I think I found the proof for the converse here.

Any help is greatly appreciated.

My attempt

If I can say that $\frac{S_n^4}{n^4} > \frac{S_n}{n}$ all the time then if the LHS goes to zero then so must the RHS. But this doesn't seem to be the case always, for example for $n=1$ I don't think I can always say that $X_1^4 > X_1$ since the only thing I know about the $X_i$'s are they have mean zero and they could take fractions as values.

Book proof

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    $\begingroup$ If $(s_n/n)^4\to 0$ then for any $\epsilon>0$ there exists a number $N(\epsilon)$ such that $\left| \frac{S_n}{n}\right|^4<\epsilon^4$ whenever $n>N(\epsilon)$. Hence, for any $\epsilon$, $\left|\frac{S_n}{n}\right|<\epsilon$ whenever $n>N(\epsilon)$. Done. $\endgroup$ – Mark Viola Oct 5 '18 at 20:56
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    $\begingroup$ Do you don't get why $(S_n/n)^4\to 0 \implies S_n/n\to 0$ ? It's because $\root{4}\of{x}$ is continuous, so $\root{4}\of{(S_n/n)^4}=|S_n/n|\to \root{4}\of{0}=0$. Finally, we get rid of absolute value (I think definition is the most straightforward for this) $\endgroup$ – Jakobian Oct 5 '18 at 20:57
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    $\begingroup$ For big enough $n$, we would have $\frac{S_n^4}{n^4}\leq \big|\frac{S_n}{n}\big|$, actually. It's because $x^4\leq |x|$ for $|x|\leq 1$. And since $\frac{S_n}{n}$ goes to $0$, it needs to eventually be less than $1$ $\endgroup$ – Jakobian Oct 5 '18 at 21:06
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    $\begingroup$ Even if this inequality doesn't imply convergence of the other sequence, we can still show that it's convergent. In fact, we don't need that inequality $\endgroup$ – Jakobian Oct 5 '18 at 21:31
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    $\begingroup$ Not necessarily. $x_n=1/n$ and $f(x)=1/x$ is what comes to mind. Though, if it's a closed interval, it should always work. In fact, if $x_n\to x$, then it should be $f(x_n)\to f(x)$ $\endgroup$ – Jakobian Oct 5 '18 at 21:50

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