How do I find the sum of the infinite geometric series #1/2# + 1 + 2 + 4 +... ?

2 Answers
Jun 5, 2018

Answer:

see below

Explanation:

for a GP the sum to infinity

#S_(oo)=a/(1-r)#

#a=" first term, "r=" the common ratio "#

only exists if

#|r|<1#

for the given GP

#1/2+ 1+2+4+..#

#r=1/(1/2)=2#

#:.|r|>1#

#=> " sum to infinity does not exist"#

Jun 9, 2018

Answer:

#S_n=1/2Sigma_(i=0)^n(2^i)->oo# as #n->oo#, i.e. the sum is indefinite.

Explanation:

You can write the sum of n first terms in the geometric series as #S_n=1/2Sigma_(i=0)^n(2^i)#
It's a well known fact that #2^n->oo# as #n->oo#. Therefore, as the individual terms go towards infinity, so must the sum.

Therefore #S_n=1/2Sigma_(i=0)^n(2^i)->oo# as #n->oo#