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Given a number $x\in (0,1)$ with decimal representation $0.a_1a_2a_3...$. Let $f(x)= 1$ iff the sequence $\sum_{i=1}^{n} {\frac{a_i}{n}}$ converges. Takes $X$ to be uniform$(0,1)$, what is the expected value of $f(X)$? Also, what can we say about the set of $x$ such that the sequence diverges?

Edit: I'm fairly sure the answer to the first question is 1. I'm mostly interested now in whether the set of divergent x has any nice properties.

davik
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  • Is it true, that law of large numbers, or something similar would tell us this converges with probability one? – davik May 13 '16 at 15:00
  • I'm guessing it's one, if it is, I'm also interested in the properties of the divergent set. – davik May 13 '16 at 15:01
  • I'd be interested in the answer too, but would it really converge with probability 1? It seems to me like it would diverge with probability 1...it's hard to make the intuition rigorous, but for any random element, roughly 1/10 of the decimals are 0, at at the other 9/10 of the decimals points it is greater than the harmonic series at the same point, if that makes sense. Not sure. – Jason Gaitonde May 13 '16 at 15:05
  • I think you may have misunderstood, I'm not dividing $a_i$ by i I'm dividing by n, so I'm finding a running average of the digits – davik May 13 '16 at 15:08
  • Oh I see, didn't realize $n$ was in the denominator...stupid me. In that case, I believe you are correct. – Jason Gaitonde May 13 '16 at 15:10
  • Actually, I think it is one, since almost all (in the measure sense) are normal, and this limit should exist for normal numbers, I think, however, the divergent set, I believe, is smaller than just the abnormal numbers. – davik May 13 '16 at 15:15
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    Relevant? http://mathoverflow.net/questions/84772/bounded-sequences-with-divergent-ces%C3%A0ro-mean – kennytm May 13 '16 at 15:53

1 Answers1

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Juse to talk about the (sub)set of diverging $x$ here.

The sequence $c_n = \frac1n\sum_{i=1}^na_i$ is known as the Cesàro mean. One way to make the mean diverge with a bounded sequence is to make it slowly diverging, e.g. let $$a_i=(-1)^{\lfloor \log_2 i \rfloor} + 1 \implies x=0.2\underbrace{00}_2\underbrace{2222}_4\underbrace{00000000}_8\underbrace{2222222222222222}_{16}\dots,$$ then we could find two subsequences with different limits $$c_{2^k-1} = \begin{cases} \frac23 & \text{$k$ is even}\\ \to \frac43 & \text{$k$ is odd}, \end{cases}$$ i.e. $c$ diverges and $f(x)=0$.

Any diverging sequence $b_i$ can be converted to supersequence $a_i=b_{T(i)}$ with $T(i)$ being logarithmic or slower which the Cesàro mean diverges.

Community
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kennytm
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