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Alfredo Holmes

Answered question

2022-03-27

Convergence of

(b_{n})

b = b_{0} . b_{1} b_{2} \dots b_{n} \dots

aznluck4u72x4

Beginner2022-03-28Added 16 answers

Step 1
You're absolutely right that the eventually you need

b_{n} = b_{n + 1} = \dots

since each

b_{n}

can only be selected from

0, 1, \dots, 9

So either you're repeating 0's or repeating one of 1 through 9's.
Lets assume for simplicity that

b_{0} = 0

If you're repeating 0's, then

b = \frac{N}{10^{n}}

for some

N < 10^{n}

and n is the last nonzero digit.
If you're repeating k's for

k \in {1 \dots 9}

, then

b = \frac{N}{10^{n}} + \frac{k}{9 \times 10^{n}}

where n is the last nonrepeating digit.
For example,

0.58777777 \dots = \frac{58}{100} + \frac{7}{900}

Note that this does not guarantee that

\frac{N}{10^{n}}

\frac{k}{9 \times 10^{n}}

can't be simplified further.
So, you could write

(b_{n})

converges if and only if there exists

n \in N, N \in N_{0}

satisfying

N < 10^{n}

, and

k \in {0, \dots 9}

such that

b - b_{0} = \frac{N}{10^{n}} + \frac{k}{9 \times 10^{n}}

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