s-i-1-2-i-2s-s-1-s-1-how-to-explain-it-and-how-to-judge-which-case-can-use-this-way-

Question Number 210788 by liuxinnan last updated on 19/Aug/24

s = \underset{i = 1}{\sum^{\infty}} 2^{i}

2 s = s - 1

s = 1 ?

h o w t o e x p l a i n i t

a n d h o w t o j u d g e w h i c h c a s e c a n u s e t h i s w a y

Answered by Berbere last updated on 19/Aug/24

w h e n Y o u h a v e e q a t i o n

t h e v a r i a b l e i s i n s e t

2 s = s - 1 h a v e s \in R . . ?

R =] - \infty, \infty [

s = \underset{i = 1}{\sum^{\infty}} 2^{i} > 2^{n}; \forall n \in N

s \to \infty \notin R

Commented by liuxinnan last updated on 20/Aug/24

t h a n k s