0
$\begingroup$

I am in need of help with the following question.

(a) Let $(d_{i})_{i=1}^{\infty}$ be a sequence of integers with $0\leq d_{i}\leq 9$. Prove that the series $\sum_{i=1}^{\infty}(d_{i}\cdot10^{-i})$ converges

(b) If the sequence $(d_{i})_{i=1}^{\infty}$ is periodic, then show that $\sum_{i=1}^{\infty}(d_{i}\cdot10^{-i})$ is a rational number. Show this number.

For part (a) I am thinking intuitively that the numbers in the series will be bounded. If $0\leq d_{i} \leq 9$ then $0 \leq d_{i}\cdot10^{-i} \leq 1$. Should I be thinking about taking the limit?

For part (b) I am unsure where to begin.

  • 1
    I won't give you the solution, but I think you might be interested in the following fraction example $\frac{1234}{9999}$.2012-03-08

2 Answers 2

1

For part a) we can bound by a geometric series.

$0 \leq\sum_{i=1}^{\infty}\frac{d_i}{10^i} \leq \sum_{i=1}^{\infty}\frac{9}{10^i}= 9\sum_{i=1}^{\infty}\left(\frac{1}{10}\right)^{i}=9\frac{1}{9}=1 $

For part b) think about each $d_i$ as being the digit in the $i$th decimal place of some real number in $[0,1]$. What does periodicity imply here?

0

A hint for b): Assume that the $d_i$ are periodic with period $m\geq1$. Then you can split up the given sum in the form $\sum_{i=1}^\infty d_i 10^{-i}= \sum_{r=0}^\infty\Bigl(\sum_{i= rm+1}^{rm +m} d_i 10^{-i}\Bigr)=\ldots\quad.$