M marcello Golden Member Aug 30, 2004 1,141 0 0 Feb 5, 2007 #1 Is the sum of (1/2)^n from 1 to infinity 1? Thanks, forgot all my math schooling. Edited title
W WA261 Diamond Member Aug 28, 2001 4,631 0 0 Feb 5, 2007 #2 (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
(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
T thesurge Golden Member Dec 11, 2004 1,745 0 0 Feb 5, 2007 #3 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
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 Lifer Feb 23, 2005 11,943 542 126 Feb 5, 2007 #4 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
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
B BigPoppa Golden Member Oct 9, 1999 1,930 0 0 Feb 6, 2007 #6 Pretty sure 1/x^r converges for all x>1 AND r>1
Evadman Administrator Emeritus<br>Elite Member Feb 18, 2001 30,990 5 81 Feb 6, 2007 #7 Sorry, I would reply but because I sqrt(-1/64) I am going to hit the sack early.
TuxDave Lifer Oct 8, 2002 10,572 3 71 Feb 6, 2007 #8 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
R RossGr Diamond Member Jan 11, 2000 3,383 1 0 Feb 6, 2007 #10 Dare I say it? This is the binary equivelent to .999... In binary .1 + .01 + .001 +... = .111... = 1
silverpig Lifer Jul 29, 2001 27,709 11 81 Feb 6, 2007 #11 Originally posted by: RossGr Dare I say it? This is the binary equivelent to .999... In binary .1 + .01 + .001 +... = .111... = 1 Click to expand... MY EARS! THE EARPLUGS DO NOTHING!!!
Originally posted by: RossGr Dare I say it? This is the binary equivelent to .999... In binary .1 + .01 + .001 +... = .111... = 1 Click to expand... MY EARS! THE EARPLUGS DO NOTHING!!!