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Infinite Series

Key Questions

  • # lim_{n rarr oo) sum_(r=0)^n a_r = c # where # c in RR #

  • A series is a sum of infinite terms, and the series is said to be divergent if its "value" is #infty#. Of course, #infty# is not a real value, and is in fact obtained via limit: you define the succession #s_n# as the sum of the first #n# terms, and study where it heads towards.

    A necessary condition for the series to converge is that the terms tend to zero. This is quite intuitive: if you add infinite terms, and those terms are always "big", your sum becomes bigger and bigger at every step, and so it must diverge. So, for example, any series which sum costants must diverge:

    #sum_{n=1}^infty 1 = infty#,

    #sum_{n=1}^infty 0.6 = infty#,

    #sum_{n=1}^infty -2 = -infty#,

    The situation is even worse if the terms you're adding grow bigger:

    #sum_{n=1}^infty n = infty#,

    #sum_{n=1}^infty n^2+3 = infty#,

    Finally, even if the terms tend to zero, some series diverge anyway: the main example is

    #sum_{n=1}^infty 1/n = infty#.

  • Answer:

    Yes.

    For example #1+1/2+1/4+1/8+1/16+... = sum_(n=0)^oo2^-n=2#

    Explanation:

    In general, if #r in (-1, 1)# and #a != 0#, then:

    #sum_(n=0)^oo ar^n = a/(1-r)#

    To see this notice that:

    #(1-r)sum_(n=0)^oo ar^n = sum_(n=0)^oo ar^n - r sum_(n=0)^oo ar^n#

    #= sum_(n=0)^oo ar^n - sum_(n=1)^oo ar^n = ar^0 = a#

    If #abs(r) > 1# then #sum_(n=0)^oo ar^n# will not converge.

    One very useful infinite series that converges for any #x in RR# is

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

    The transcendental number #e# is

    #exp(1) = 1+1/(1!)+1/(2!)+1/(3!)+... ~= 2.71828182844#

    In fact, you can show that #exp(x) = e^x#

Questions