A periodic decimal expansion sequence Let (x_n)_{n \in \mathbb{N}} be a

Reginald Owens

Reginald Owens

Answered question

2022-03-03

A periodic decimal expansion sequence
Let (xn)nN be a sequence with the following terms.
x1=0, 1
x2=0, 13
x3=0, 131
x4=0, 1313
x5=0, 13131
and son on. Prove limnxn=1399

Answer & Explanation

lucratifar1

lucratifar1

Beginner2022-03-04Added 5 answers

Step 1
I'll start with an independent proof that
1399=0.1.3.:
Let
 x=0.1.3..
Then
100x=13.1.3.
99x=100xx=13.1.3.0.1.3.=13
x=1399=0.1.3..
1399=0.1.3.=0.13131313 .
Hence for all natural numbers k, we have:
0<1399xk<110k.
Take limits as k of the above inequality and quote the Squeeze Theorem.
An alternative method would be to prove that the sequence is Cauchy which is similar to the above method in fact, and then use the fact that every real Cauchy sequence (with the usual metric) is convergent.

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