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Math Help - limits and geometric series

  1. #1
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    limits and geometric series

    A) suppose that |r| < 1 find the limit:
    Lim n->(infinity) (1+r+r^2+....+r^n)
    B) If we let the infinite repeating decimal .9999 ...... stand for the limit
    Lim n->(infinity) ( 9/10 + 9/(10^2) + ..... + 9/(10^n) ), show that .9999..... = 1
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by luckyc1423 View Post
    B) If we let the infinite repeating decimal .9999 ...... stand for the limit
    Lim n->(infinity) ( 9/10 + 9/(10^2) + ..... + 9/(10^n) ), show that .9999..... = 1
    let 0.999999.... = lim(n--> oo) (9/10 + 9/(10^2) + ..... + 9/(10^n) )
    .....................= lim(n--> oo) 9/10 (1 + 1/10 + 1/(10^2) + ....)
    .....................= lim(n--> oo) 9/10 * lim(n--> oo) (1 + 1/10 + 1/(10^2) + ....)
    .....................= 9/10 *lim(n--> oo) (1 + 1/10 + 1/(10^2) + ....)

    Note that (1 + 1/10 + 1/(10^2) + ....) is a geometric series with a = 1 and r = 1/10 (which means |r| < 1). so it's infinite sum is given by a/(1 - r)

    ......................= 9/10 * 1/(1 - 1/10) = 9/10 * 1/(9/10) = 9/10 * 10/9 = 1
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by luckyc1423 View Post
    A) suppose that |r| < 1 find the limit:
    Lim n->(infinity) (1+r+r^2+....+r^n)
    note that 1+r+r^2+....+r^n is a geometric series with a = 1 and r = r, so it's infinite sume is given by a/(1 - r) when |r|< 1

    so Lim n->(infinity) (1+r+r^2+....+r^n) = 1/(1 - r)
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