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Let $x$ be a real number in $[0,1)$, with decimal expansion $$ x = 0.d_1 d_2 d_3 \cdots d_i \cdots \;. $$ If the decimal expansion is finite, ending at $d_i$, then extend with zeros: $d_k = 0$ for all $k > i$. Define a sequence $x_k^R$ by digit reversals, as follows: \begin{eqnarray} x_1^R & = & 0.d_1 \\ x_2^R & = & 0.d_2 d_1 \\ x_3^R & = & 0.d_3 d_2 d_1 \\ x_4^R & = & 0.d_4 d_3 d_2 d_1 \\ & \cdots &\\ x_k^R & = & 0.d_k d_{k-1} \cdots d_3 d_2 d_1\\ & \cdots & \end{eqnarray} Finally, define $x^R = \lim_{k\to\infty} x_k^R$, when that limit exists.

Q. For which $x$ does the limit exist? In particular, must $x$ be rational for the limit $x^R$ to exist? If not, what are some irrationals with limits?

If the decimal expansion of $x$ is finite, then the extension by zeros leads to $\lim_{k\to\infty} x_k^R = 0$.

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    $\begingroup$ It seems pretty clear that the limit can exist if and only if $d_k$ is eventually constant, doesn't it? (This includes the case of being eventually zero.) $\endgroup$
    – MPW
    Mar 14 '16 at 12:28
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    $\begingroup$ It could also be eventually 9, so you may have two choices. $\endgroup$ Mar 14 '16 at 15:58
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    $\begingroup$ Note that even if $d_k$ is not eventually constant, the set $x_k$ can still have a finite number of accumulation points; in fact, for every rational number, it will have as many accumulation points as there are digits that repeat in the decimal expansion of $x$. $\endgroup$ Mar 14 '16 at 18:12
  • $\begingroup$ @MichaelSeifert: I do wonder about a sequence of trailing digits that is not quite constant, but increasingly nearly constant... That was behind my question, and perhaps that is your point. $\endgroup$ Mar 14 '16 at 21:16
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The limit exists precisely when the sequence of digits is eventually constant. If it is eventually constantly $d$, the limit is $\frac{d}9$.

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  • $\begingroup$ Suppose the sequence is not eventually constant, but rather an increasingly long sequence of constant digits, but with an infrequent blip. $\endgroup$ Mar 14 '16 at 21:18
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    $\begingroup$ @Joseph: Then the sequence of reversals won’t converge; see Crostul’s proof above. With more information, we can say more. Suppose, for instance, that the mostly constant digit is $d$, and the blips are single digits; then there is some digit $a$ that occurs infinitely often as a blip, the sequence of reversals has one subsequence that converges to $\frac{d}9$ and another that converges to $\frac{a}{10}+\frac{d}{90}$. In particular, it clearly does not converge. $\endgroup$ Mar 14 '16 at 21:27
  • $\begingroup$ @Joseph: You’re welcome. $\endgroup$ Mar 14 '16 at 21:35
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Suppose that $\lim_{k \to \infty} x^R_k = x^R$ exists. Then, for $k$ large enough, the first digit of $x^R_k$ is equal to the first digit of $x^R$, say $a$.

Hence, $d_k = d_{k+1} = \dots = a$ and $x^R = 0.aaaa \dots = a/9$.

So, the number $x$ must have a decimal expansion where eventually only the digit $a$ appears.

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    $\begingroup$ ...and, for completeness, that implies that $x$ is rational. $\endgroup$
    – psmears
    Mar 14 '16 at 15:57
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    $\begingroup$ but not ALL rationals: only those equal to m + 1/9*k*(10^-n) for m, k, n positive integer. $\endgroup$ Mar 14 '16 at 16:27

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