An alternating decimal sequence: Does its average have a limit?

Question

Define a sequence of decimals $x_n$ by alternating the digits $1,2,\ldots,n$ left and right, as follows: $$x_1 = .1$$ $$x_2 = .21$$ $$x_3 = .213$$ $$x_4 = .4213$$ $$x_5 = .42135$$ $$x_6 = .642135$$ $$\cdots$$ $$x_{14}=.1412108642135791113$$ $$\cdots$$ Define $a_n$ as the average of the sequence up to $x_n$: $$a_n = (x_1+x_2+ \cdots x_n)/n$$

Q. Does $a_n$ approach a limit as $n\to \infty$?

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Addendum. Resolved by Thomas Andrews: the sequence has no limit! However, Paul Hurst conjectures that $\lim \sup_{n \to \infty} = 0.55$.

The same question can be asked in any base $b$. Especially interesting is $b=2$, e.g., $x_3=.10111$ (i.e.,$10{-}1{-}11$), $x_7=.11010010111101111$ (i.e.,$110{-}100{-}x_3{-}101{-}111$), etc.

Answer 1

There is no limit. Assume that the limit is $L\leq 0.55$. Then let $N=6\cdot 10^k$ and $M=7\cdot 10^k$, and note that $$a_M = \frac{6\cdot10^k a_N + \sum_{i=N+1}^M x_i}{7\cdot 10^k}$$

Now, $\sum_{i=N+1}^M x_i \geq 0.6\cdot 10^k\geq (L+0.05)\cdot 10^k$. So we have:

$$a_M \geq \frac{6a_N+L+0.05}{7}= \frac{6a_N+L}{7} + \frac{0.05}{7}$$

Letting $k$ got to infinity, this would mean $L \geq L + \frac{0.05}{7},$ which is a contradiction.

The case $L\geq 0.55$ also isn't possible. A similar argument, using $N=4\cdot 10^k$ and $M=5\cdot 10^k$ then works.

Basically, while the values $x_i$ might be relatively uniform in the interval $[0.1,1]$, they come in huge blocks that have the same leading digit, which causes the sequence $a_N$ to be not Cauchy.

It seems likely that $\lim_{k\to\infty} a_{10^k} = 0.55$, though.

Proof without cases that $a_n$ isn't Cauchy.

Lemma: $a_{10^k}\geq 0.45$ for $k>1$.

Proof: $$a_{10^k} = \frac{10^{k-1}a_{10^{k-1}} + \sum_{i=1}^9\sum_{j=1}^{10^{k-1}} x_{i 10^{k-1} + j}}{10^k}$$

Now, $x_{i10^{k-1}+j}\geq \frac{i}{10}$. So this means:

$$a_{10^k}\geq \frac{a_{10^{k-1}}}{10} + \frac{1}{100}\sum_{i=1}^9 i = \frac{a_{10^{k-1}}}{10} + 0.45\geq 0.45$$

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You can use this lemma with my original argument to show that $a_{2\cdot 10^k} < a_{10^k}-\epsilon$ for an $\epsilon>0$ independent of $k$, and thus that $a_n$ is not Cauchy.

$$\begin{align}a_{2\cdot 10^k} &= \frac {10^ka_{10^k}+\sum_{i=1}^{10^k}x_{10^k+i}}{2\cdot 10^k}\\&<\frac{1}{2}\left(a_{10^k}+ \frac{2}{10}\right) \\&< a_{10^k} -\left(\frac{1}{2}a_{10^k}-\frac{1}{10}\right) \\&< a_{10^k}-0.1 \end{align}$$

The last step by the Lemma, the first step by $x_{10^k+i}<0.2$.

Answer 2

I'm not 100% sure on this solution. But it looks like to me that in each digit spot you're going to end up with a uniform spread of the integers from 0 to 9. So you're going to have an average of 4.5 in each digit spot. $$\frac{4.5}{10} + \frac{4.5}{10^2} + \frac{4.5}{10^3} +\cdots = \frac{4.5}{10}(1+\frac{1}{10} + \frac{1}{10^2} + \cdots) = 0.5$$

Edited: But the digit in the tenths slot is always between 1 and 9, which average out to 0.5, so that would give $$\frac{5}{10} + \frac{4.5}{10^2} + \frac{4.5}{10^3} +\cdots = 0.55$$ But I'm not sure if this converges because you get lots of consecutive digits in the first slot.

I think Thomas Andrews just proved it diverges. However, I would conjecture that the $\limsup$ is 0.55.