Sum of (1/2^n)

[marcello]

Is the sum of (1/2)^n from 1 to infinity 1? Thanks, forgot all my math schooling.

Edited title

 

[WA261]

(n + 1)2 = n2 + O(n)
(n+O(n^{1/2}))(n + O(\log\,n))^2 = n^3 + O(n^{5/2})
nO(1) = O(en)

wait..errr..that is just n->inf..nm

 

[thesurge]

The sum of an infinite series in geometric progression is:

a_1/(1-r) with absolute convergence when |r|<1 (r is common ratio, or a_{n+1}/a_n)

so

(1/2^{2})/(1-[1/(2^(n+1))]/[1/(2^(n))])=1

 

[Cerpin Taxt]

Yes, pretty sure the sum is 1. If you would've said from n = 0 to infinity, then the sum would be 2, since (1/2)^0 = 1

 

[marcello]

Sweet, I thought so. Thanks!

 

[BigPoppa]

Pretty sure 1/x^r converges for all x>1 AND r>1

 

[Evadman]

Sorry, I would reply but because I sqrt(-1/64) I am going to hit the sack early.

 

[TuxDave]

1/2+1/4+1/8+1/16+...=x
1/2(1+1/2(1+1/2(... = x
1/2(1+x) = x
1+x=2x
x=1

 

[chuckywang]

1, it's just a infinite geometric sum.

 

[RossGr]

Dare I say it?





This is the binary equivelent to .999...

In binary

.1 + .01 + .001 +... = .111... = 1

 

[silverpig]

Originally posted by: RossGr
Dare I say it?





This is the binary equivelent to .999...

In binary

.1 + .01 + .001 +... = .111... = 1

MY EARS! THE EARPLUGS DO NOTHING!!!