Infinite Series

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What is a Series
12:24 — by patrickJMT

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Key Questions

  • It is very tough to answer such a general question, but I will give it a shot. Infinite series allow us to add up infinitely many terms, so it is suitable for representing something that keeps on going forever; for example, a geometric series can be used to find a fraction equivalent to any given repeating decimal such as:

    #3.333...#

    by splitting into individual decimals,

    #=3+0.3+0.03+0.003+cdots#

    by rewriting into a form of geometric series,

    #=3+3(1/10)+3(1/10)^2+3(1/10)^3+cdots#

    by using the formula for the sum of geometric series,

    #=3/{1-1/10}=10/3#

    The knowledge of geometric series helped us find the fraction fairly easily. I hope that this was helpful.

  • Short answer:
    The key step is to determine if the term a_n in the infinite series is one side bounded or not (in the limit).

    Long answer:
    Assume that the infinite series is given by #{a_n}_{n=1}^{infty}#, i.e., #a_n# is the n-th term in the series.
    1. First step is to check if the term is bounded by a positive finite value for all n
    2. Second, check if the term reaches a finite constant in the limit i.e.,
    #lim_{n -> infty} a_n = c #
    More precisely, for a given #epsilon > 0# (a very small positive real number) there exists a integer N such that
    #|a_n - c| < epsilon # if # n > N #.
    This basically means #a_n# approaches c as n increases.

    If a series satisfies this test, then it is convergent.
    For example #a_n = (1+2/n)/(2+3/n^2) # is convergent because, for all n, terms a_n are smaller than or equal to #c = 1/2#.
    If this test fails, ie., there is no finite term c by which #a_n# is bounded then the series is divergent.
    For instance, #a_n = e^n/n^2# is divergent as its values increases without bound.
    Finally if the series is said to oscillate if the value fluctuates between two extremes.
    eg. #a_n = (-1)^n#.
    The values are -1 and 1 for odd and even n's respectively.

    Reference:
    Chapter-section 5.1:
    A. Khuri, Advanced calculus with applications in statistics, 2nd ed, Wiley, 2003, pp. 132 - 134.

  • I believe that it is the same as an alternating series. If that is the case, then an oscillating series is a series of the form:

    #sum_{n=0}^infty (-1)^n b_n#, where #b_n ge 0#.

    For example, the alternating harmonic series

    #sum_{n=1}^infty{(-1)^n}/n#

    is a convergent alternating series.

  • Here is an example of a telescoping series
    #sum_{n=1}^infty(1/n-1/{n+1})#
    #=(1/1-1/2)+(1/2-1/3)+(1/3-1/4)+cdots#
    As you can see above, terms are shifted with some overlapping terms, which reminds us of a telescope. In order to find the sum, we will its partial sum #S_n# first.
    #S_n=(1/1-1/2)+(1/2-1/3)+cdots+(1/n-1/{n+1})#
    by cancelling the overlapping terms,
    #=1-1/{n+1}#
    Hence, the sume of the infinite series can be found by
    #sum_{n=1}^infty(1/n-1/{n+1})=lim_{n to infty}S_n=lim_{n to infty}(1-1/{n+1})=1#

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Tests of Convergence / Divergence