Convergence of Geometric Series
Key Questions

Answer:
A geometric series of geometric sequence
#u_n= u_1 * r^(n1)# 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^(n1)# And a geometric series can be written in several forms :
#sum_(n=1)^(+oo)u_n = sum_(n=1)^(+oo)u_1*r^(n1) = u_1sum_(n=1)^(+oo)r^(n1)# #= u_1*lim_(n>+oo)(r^(11) + r^(21) + r^(31) + ... + r^(n1))# Let
#r_n = r^(11) + r^(21) + r^(31) + ... + r^(n1)# Let's calculate
#r_n  r*r_n# :#r_n  r*r_n = r^(11)  r^(21) + r^(21)  r^(31) + r^(31) + ...  r^(n1) + r^(n1)  r^n = r^(11)  r^n# #r_n(1r) = r^(11)  r^n = 1  r^n# #r_n = (1  r^n)/(1r)# Therefore, the geometric series can be written as :
#u_1sum_(n=1)^(+oo)r^(n1) = u_1*lim_(n>+oo)((1  r^n)/(1r))# Thus, the geometric series converges only if the series
#sum_(n=1)^(+oo)r^(n1)# converges; in other words, if#lim_(n>+oo)((1  r^n)/(1r))# exists.
If r > 1 :
#lim_(n>+oo)((1  r^n)/(1r)) = oo# 
If r < 1 :
#lim_(n>+oo)((1  r^n)/(1r)) = 1/(1r)# .
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# . 

Answer:
Here are some examples:
#1 + 1/2 + 1/4 + 1/8 + 1/16 +...# #1  1 + 1  1 + 1  1 +...# #1 + 2 + 4 + 8 + 16 +...# Explanation:
All geometric series are of the form
#sum_(i=0)^oo ar^i# where#a# is the initial term of the series and#r# the ratio between consecutive terms.In the three examples above, we have:
#a = 1# ,#r = 1/2# #sum_(i=0)^oo ar^i = 2# #a = 1# ,#r = 1# #sum_(i=0)^oo ar^i# does not converge  it alternates between#0# and#1# as each term is added.#a = 1# ,#r = 2# #sum_(i=0)^n ar^i > oo# as#n>oo# The geometric series
#sum_(i=0)^oo ar^i# only converges in the following cases:(1)
#a = 0# #sum_(i=0)^oo ar^i = 0# (2)
#abs(r) < 1# #sum_(i=0)^oo ar^i = a/(1r)# 
#a+ax+ax^{2}+ax^{3}+cdots# , which converges to#a/(1x)# when#x<1# . More generally, you could also write#a+a(xc)+a(xc)^{2}+a(xc)^{3}+cdots# , which converges to#a/(1(xc))=a/((1+c)x)# when#xc < 1# .