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Strategies to Test an Infinite Series for Convergence

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M14: Strategies to test series for convergence

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Key Questions

  • There is no general method of determining the test you should use to check the convergence of a series.

    • For series where the general term has exponents of #n#, it's useful to use the root test (also known as Cauchy's test).
      Example 1: Power Series
      The definition of the convergence radius of the of a power series comes from the Cauchy test (however, the actual computation is usually done with the following test).

    • Generally, the computation of the ratio test (also known as d'Alebert's test) is easier than the computation of the root test.
      Example 2: Inverse Factorial
      For the series #sum_(n=1)^(oo) 1/(n!)# the d'Alembert's test gives us:
      #lim_(n to oo) |1/((n+1)!)|/|1/(n!)| = lim_(n to oo) |n!|/(|(n+1)!|) = lim_(n to oo) |(n!)/((n+1)n!)| = lim_(n to oo) |1/(n+1)| = 0#
      So the series is convergent.

    • If you know the result of the improper integral of the function #f(x)# such that #f(n)=a_n#, where #a_n# is the general term of the series being analyzed, then it might be a good idea to use the integral test.
      Example 3: A proof for the Harmonic Series.
      Knowing that the improper integral #int_1^(oo) 1/x dx# is divergent (it's easy to check) implies that the harmonic series #sum_(n=1)^(oo) 1/(n)# diverges.

    • Comparision tests are only useful if you know an appropriate series to compare the one you're analyzing to. However, they can be very powerful.
      Example 4: Hyperharmonic Series
      The series of the form #sum_(n=1)^(oo) 1/(n^p)# are called hyperharmonic series or #p#-series. If you can show that the series #sum_(n=1)^(oo) 1/(n^(1+epsilon))# converges, for some small, positive value of #epsilon#, than any #p#-series such that #p>1 + epsilon# converges.

  • By pulling out the negative sign so that #a_n ge0#, let

    #-sum_{n=1}^inftya_n=-sum_{n=1}^infty{n^2+2^n}/{e^{n+1}-1}#.

    I would use Limit Comparison Test since we can make a ball-park estimate of the series by only looking at the dominant terms on the numerator and the denominator. This series can be compared to

    #sum_{n=1}^inftyb_n=sum_{n=1}^infty{2^n}/{e^{n+1}}=sum_{n=1}^infty1/e(2/e)^n#,

    which is a convergent geometric series with #|r|=|2/e|<1#.

    Let us make sure that they are comparable.

    #lim_{n to infty}{a_n}/{b_n}=lim_{n to infty}{{n^2+2^n}/{e^{n+1}-1}}/{{2^n}/{e^{n+1}}}#

    #=lim_{n to infty}{{n^2}/{2^n}+1}/{1-1/e^{n+1}}={0+1}/{1-0}=1 < infty#

    (Note: #lim_{n to infty}{n^2}/2^n=0# by applying l'Hopital's Rule twice.)

    So, #sum_{n=1}^infty a_n# converges by Limit Comparison Test.

    Hence, #sum_{n=1}^infty{n^2+2^n}/{1-e^{n+1}}# also converges since negation does not affect the convergence of the series.


    I hope that this was helpful.

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