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For any sequence of RVs $\{X_n\}$, show that

$\max_{1\le k\le n}|X_k|\to 0$ in probability $\Rightarrow n^{-1}S_n\to 0$ in probability, where $S_n=\sum_{k=1}^nX_k, k=1,2,\ldots$

When thinking of convergence in probability, my instinct was to use the Weak Law of Large Numbers. However, I don't know how to get into the proper form. The max is confusing me. Maybe want to show that this condition implies that $\{X_n\}$ has a finite mean of $0$? Not sure if that applies in the case of non-iid.

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    I don't understand how $\max_{k=1}^n|X_k|$ could converge to zero in probability except in trivial cases (It's increasing). Are you sure you didn't mean $\sup_{k\ge n}|X_k|$ or something like that? The theorem as stated is true, but seems pretty silly since only the zero sequence would qualify. – 2017-01-20
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    I'm sure. The theorem is definitely using max. – 2017-01-20
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    Why does it say there's a deleted post here? – 2017-01-20
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    Proper notation is not {$X_n$} but $\{X_n\}$. I edited accordingly. – 2017-01-20
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    Sorry I'm trying to learn MathJax on the fly here. – 2017-01-20
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    @user408688 : An answer was posted 13 minutes ago and has since been deleted. – 2017-01-20
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    @MichaelHardy Why? Was it wrong? – 2017-01-20
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    @user408688 Do you understand what I'm saying? Am I missing something? – 2017-01-20
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    @user408688 : It was deleted by the person who posted it three minutes after it was posted. – 2017-01-20
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    @spaceisdarkgreen I get what you mean but I know for a fact the question is written correctly. Does that mean I only need to consider the zero sequence? I don't think it was supposed to be that simple. – 2017-01-20
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    You could use triangle/max property to say that $\frac{|S_n|}{n} <\frac{1}{n}\sum_{j=1}^n \max_{k=1}^n(|X_k|) = \max_{k=1}^n|X_k|$. But again, the only thing for which the RHS is zero is the zero sequence. . – 2017-01-20

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