Direct Comparison Test for Convergence of an Infinite Series

Key Questions

  • By Comparison Test, we can conclude that the series
    sum_{n=1}^infty{9^n}/{3+10^n}n=19n3+10n converges.

    Let us look at some details.
    For all n geq 1n1,
    9^n/{3+10^n} leq 9^n/10^n=(9/10)^n9n3+10n9n10n=(910)n

    By Geometric Series Test,
    sum_{n=1}^infty(9/10)^nn=1(910)n converges since |r|=9/10<1|r|=910<1

    Hence, by Comparison Test,
    sum_{n=1}^infty{9^n}/{3+10^n}n=19n3+10n also converges.

  • If an improper integral diverges, then we probably do not want to wast time trying to find its value.

    Let us assume that we already know:

    int_1^infty1/x dx=infty11xdx=

    Let us look examine this uglier improper integral.

    int_1^infty{4x^2+5x+8}/{3x^3-x-1} dx14x2+5x+83x3x1dx

    By making the numerator smaller and the denominator bigger,

    {4x^2+5x+8}/{3x^3-x-1} ge {3x^2}/{3x^3}=1/x4x2+5x+83x3x13x23x3=1x

    By Comparison Test, we may conclude that

    int_1^infty{4x^2+5x+8}/{3x^3-x-1} dx14x2+5x+83x3x1dx diverges.

    (Intuitively, if the smaller integral diverges, then the larger one has no chance to converge.)

    I hope that this was helpful.

  • If you are trying determine the conergence of sum{a_n}{an}, then you can compare with sum b_nbn whose convergence is known.

    If 0 leq a_n leq b_n0anbn and sum b_nbn converges, then sum a_nan also converges.
    If a_n geq b_n geq 0anbn0 and sum b_nbn diverges, then sum a_nan also diverges.

    This test is very intuitive since all it is saying is that if the larger series comverges, then the smaller series also converges, and if the smaller series diverges, then the larger series diverges.

Questions