Convergence of Geometric Series

Key Questions

  • How can I tell whether a geometric series converges?

    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#.

    Shura · 2 · Jul 2 2015
  • What are some examples of infinite geometric series?

    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/(1-r)#

    George C. · 1 · Jul 13 2015
  • What is the geometric power series?

    #a+ax+ax^{2}+ax^{3}+cdots#, which converges to #a/(1-x)# when #|x|<1#. More generally, you could also write #a+a(x-c)+a(x-c)^{2}+a(x-c)^{3}+cdots#, which converges to #a/(1-(x-c))=a/((1+c)-x)# when #|x-c| < 1#.

    Bill K. · 1 · Dec 7 2015

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