How can I tell whether a geometric series converges?

1 Answer
Jul 2, 2015

Answer:

A geometric series of geometric sequence #u_n= u_1 * r^(n-1)# converges only if the absolute value of the common factor #r# of the sequence is strictly inferior to #1#; in other words, if #|r|<1#.

Explanation:

The standard form of a geometric sequence is :

#u_n = u_1 * r^(n-1)#

And a geometric series can be written in several forms :

#sum_(n=1)^(+oo)u_n = sum_(n=1)^(+oo)u_1*r^(n-1) = u_1sum_(n=1)^(+oo)r^(n-1)#

#= u_1*lim_(n->+oo)(r^(1-1) + r^(2-1) + r^(3-1) + ... + r^(n-1))#

Let #r_n = r^(1-1) + r^(2-1) + r^(3-1) + ... + r^(n-1)#

Let's calculate #r_n - r*r_n# :

#r_n - r*r_n = r^(1-1) - r^(2-1) + r^(2-1) - r^(3-1) + r^(3-1) + ... - r^(n-1) + r^(n-1) - r^n = r^(1-1) - r^n#

#r_n(1-r) = r^(1-1) - r^n = 1 - r^n#

#r_n = (1 - r^n)/(1-r)#

Therefore, the geometric series can be written as :

#u_1sum_(n=1)^(+oo)r^(n-1) = u_1*lim_(n->+oo)((1 - r^n)/(1-r))#

Thus, the geometric series converges only if the series #sum_(n=1)^(+oo)r^(n-1)# converges; in other words, if #lim_(n->+oo)((1 - r^n)/(1-r))# exists.

  • If |r| > 1 : #lim_(n->+oo)((1 - r^n)/(1-r)) = oo#

  • If |r| < 1 : #lim_(n->+oo)((1 - r^n)/(1-r)) = 1/(1-r)#.

Therefore, the geometric series of geometric sequence #u_n# converges only if the absolute value of the common factor #r# of the sequence is strictly inferior to #1#.