# For what values of $a$ will the following sequence converge?

$a \in \mathbb R$ has the decimal expansion $a = a_0.a_1a_2a_3 \ldots a_n \ldots$

Find all values for $a$ for which the sequence $\{a_n\}_{n=1}^{\infty}$ converges.

I rule out irrationals first, because if they can't be represented with integer numerator and denominator, they can't have convergent decimal expansions. But how do I separate the rationals, say differ between say $1/7$ and $1/3$?

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As you have observed since $a_k$'s are digits from $0$ to $9$, it will converge only when there exists an $N$ such that $a_n = d \in \{0,1,2,3,4,5,6,7,8,9\}$ for all $n>N$.

(To see this, choose say $\epsilon = 0.1$. Since $a_n$ converges, this means there exists $N(\epsilon)$, such that $\vert a_{n+1} - a_n \vert < \epsilon$ for all $n > N$. But $a_n, a_{n+1} \in \{0,1,2,\ldots,9\}$. Hence, $a_{n+1} = a_n$.)

Hence, $$a = a_0.a_1a_2\ldots a_N dddddd \ldots$$ $$10^Na = a_0 a_1 \ldots a_N . dddd\ldots$$ $$10^{N+1}a = a_0 a_1 \ldots a_N d. dddd\ldots$$ Hence, $$10^{N+1}a - 10^Na = (a_0 a_1 \ldots a_N d - a_0 a_1 \ldots a_N) = M$$ $$a = \dfrac{M}{9 \times 10^N}$$ i.e. in the simplest form ($\gcd$ of numerator and denominator is $1$) the numbers are of the form $$\dfrac{M}{2^{n_1}3^{n_2}5^{n_3}}$$ where $n_1,n_3 \in \mathbb{N}$, $n_2 \in \{0,1,2\}$ and $M \in \mathbb{Z}$ such that $\gcd(M,2^{n_1}3^{n_2}5^{n_3}) = 1$.

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Hint: This is a sequence of integers, mostly between $0$ and $9$. It should not be hard to show that it converges iff it is eventually constant.

What real numbers have an evenually constant decimal expansion? Maybe treat eventually $0$ and eventually $9$ separately (you don't need to), and note that $\dfrac{1}{9}=0.1111111\dots$.

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this is what i'm having difficulty with. rule out the irrationals, so given a rational $a/ b$ is there a way to say this will eventually go constant? – pad Oct 30 '12 at 0:03
No, a general rational will not go eventually constant, almost all of them will not. For example, ask your calculator about $1/7$, or $1/11$. Rationals have decimal expansion which is eventually periodic. – André Nicolas Oct 30 '12 at 0:07
Take a rational which is eventually constant. By mltiplying by a suitable power of $10$, say $10^k$, we get something of form $0.ddddd\dots$. This part is $\dfrac{d}{9}$. So our number is $\dfrac{1}{10^k}(N+\dfrac{d}{9})$. Bring to a common denominator. We get that our original number has shape $\dfrac{A}{9\times 10^k}$. – André Nicolas Oct 30 '12 at 0:10

A sequence of digits converges if and only if it is eventually constant. If $a\in\Bbb R$, the sequence of digits in the decimal expansion of $a$ is eventually constant if and only if there is some $n\in\Bbb N$ such that the fractional part of $10^na$ is $0$ or $0.ddd\dots$ for some $d\in\{1,\dots,9\}$. Since $0.ddd\dots=\frac{d}9$, you’re looking for real numbers $a$ such that $10^na=\pm\left(m+\frac{d}9\right)$ for some some $m,n\in\Bbb N$ and $d\in\{0,1,\dots,8\}$. (The case $d=9$ is covered by the case $d=0$.) From there you should be able to work out a nice description of these real numbers without too much trouble.

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