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Is every infinite decimal sequence convergent? For example, would this sequence $x= 0.12112211122211112222\ldots$ be considered convergent?

Dr. Mathva
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    $x=0.12112211122211112222....$ is not a sequence. $(0.1,0.12,0.1211,0.121122,0.121122111,...)$ is, and it is convergent. – player3236 Oct 07 '20 at 20:51

3 Answers3

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Yes, so long as there are only finitely many terms prior to the decimal. For any given number $n\in (0,1)$ with digits $a_i$ to the right of the decimal you can write it as:

$$n=\sum_{i=1}^\infty \frac{a_i}{10^i},$$

which converges absolutely, since $n\leq \sum_{i=0}^\infty 10^{-i}=10/9$.

Alex R.
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    I think the series $\sum_{i=0}^{\infty}10^{-i}$ should be $\sum_{i=0}^{\infty}9\cdot 10^{-i}$ for this argument to work. – Alann Rosas Oct 07 '20 at 21:01
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Yes.

Let $x_n$ be the rational number obtained by cutting after $n$ decimals. Then the sequence of the $x_n$ is non-decreasing and is bounded form above by (integer part $+1$), hence is convergent. The real number that is the limit is what we say the given decimal sequence represents.

(Note that this also implies once more that $0.9999\ldots=1$, for example)

Hagen von Eitzen
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Yes, every decimal sequence is convergent. As pointed out, you can look at the sequence of partial decimal expansions, (0.a1,0.a1a2,...). To see that it is convergent, recall that in the reals, every Cauchy sequence is convergent. The sequence is clearly Cauchy, because the difference between the n^th term and the (n+1)^st term is less than (1/10)^n, so the terms get as close to one another as you might like, which is exactly the definition of Cauchy.

mwalth
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