Infinite Series

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M10-9: Series; generalities

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1 of 2 videos by Calculus V.

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.

  • Answer:

    See below

    Explanation:

    There are different types of series, to what use different methods of evaluating

    For example a converging geometric series:

    #a+ ar + ar^2 + ar^3 + ... + ar^k = sum_(n=1) ^(k) ar^(n-1) #

    where #sum_(n=1) ^(k) ar^(n-1) = (a(1-r^k)) / (1-r) #

    Assuming #|r| < 1 # we can let #k to oo # for infinite series to be evaluated ...

    #lim_(k to oo ) sum_(n=1) ^(k) ar^(n-1) = lim_(k to oo ) ( a(1-r^k) )/(1-k) #

    as #k to oo # , #r^k to 0 # as #|r|< 1 #

    #=> sum_(n=1) ^(oo) ar^(n-1) = a/(1-r) #

    but there are other series what can be approached with tricks!

    Take # 1/6 + 1/12 + 1/20 + 1/30 + ... #

    After consideration we can recognise this is the same as...

    # (1/2 - 1/3 ) + ( 1/3 - 1/4) + (1/4 - 1/5 ) + ... #

    # (1/2 cancel(- 1/3) ) + (cancel( 1/3) cancel(- 1/4)) + (cancel(1/4) cancel(- 1/5) ) + ... #

    #=1/2 #

    There are also other infinite series that you can remember, and may be able to prove, a like:

    #e^x = 1 + x + x^2 /(2!) + x^3 / (3!) + ... = sum_(n=0) ^oo x^n / (n!) #

    There are many others, where there insist one set way of computing infinite series, there are many!

  • 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#

Questions

  • moutar answered · 2 years ago