How do I show that the sum of a convergent geometrical series 1+r+r^2+... > 1/2?

I understand how to do it for fixed numbers, but i dont know how to find the r value in this case. Thanks for the help!

1 Answer
Mar 4, 2018

See below.

Explanation:

Calling #S_n(r)=sum_(k=0)^n r^k =( r^(n+1)-1)/(r-1)#

now if #lim_(n->oo)S_n # converges then #abs r < 1# and then

#S_(oo)(r) = lim_(n->oo)S_n(r) = 1/(1-r)#

so if we need

#S_(oo)(r) > 1/2 rArr 1/(1-r) > 1/2 rArr r > -1# which means that for #r in (-1,1)# we have #S_(oo)(r) > 1/2#