+0  
 
0
49
1
avatar+12 

How can we prove that \(\frac{1}{b-1}=0.\overline{1}_b\) is always true?

 Feb 23, 2020
 #1
avatar+6180 
0

\(0.\overline{1}_b = \\ \sum \limits_{k=1}^\infty \dfrac{1}{b^k} \\ \text{let $u = \dfrac 1 b$}\\ \sum \limits_{k=1}^\infty \dfrac{1}{b^k} = \sum \limits_{k=1}^\infty u^k = \\ \dfrac{u}{1-u} = \\ \dfrac{\frac 1 b}{1-\frac 1 b} = \\ \dfrac{1}{b-1} \)

.
 Feb 24, 2020

27 Online Users

avatar