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Math Help - Show sequence convergence

  1. #1
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    Show sequence convergence

    1. Let (x_n) be a sequence with x_1 = 1 and x_n = x_{n-1} - \frac{(-1)^n}{n}. Prove convergence.

    I tried to prove this is cauchy, but couldn't make much progress.

    2. Let (s_n) be a sequence with s_1 = 3, s_{n+1} = \frac{2}{3}s_n + \frac{4}{3s_n}. Prove convergence and evaluate the limit.

    I know the limit is 2, but I don't understand how to show this using an \epsilon-proof.
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  2. #2
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by chrischen88 View Post
    1. Let (x_n) be a sequence with x_1 = 1 and x_n = x_{n-1} - \frac{(-1)^n}{n}. Prove convergence.

    I tried to prove this is cauchy, but couldn't make much progress.

    2. Let (s_n) be a sequence with s_1 = 3, s_{n+1} = \frac{2}{3}s_n + \frac{4}{3s_n}. Prove convergence and evaluate the limit.

    I know the limit is 2, but I don't understand how to show this using an \epsilon-proof.
    1. Write out the first few terms:

    x_1=1
    x_2=1-\frac{1}{2}
    x_3=1-\frac{1}{2}+\frac{1}{3}
    ...
    x_n=\sum_{k=1}^n\frac{(-1)^{n+1}}{n}

    The alternating series test says that this converges. (It happens to converge to \ln2, but that's neither here nor there.)

    2. You don't need an \epsilon-proof. I would try to prove that it's monotonically decreasing, and convergence would follow from there (because it's obviously bounded below by 0). You should be able to do it by induction, though I won't go into the details, because it's likely to be tedious. Once you know that it converges, finding the number to which it converges is easy:

    \lim_{n\to\infty}s_{n+1}=\lim_{n\to\infty}\left(\f  rac{2}{3}s_n+\frac{4}{3s_n}\right)\implies s=\frac{2}{3}s+\frac{4}{3s}\implies\frac{s^2-4}{3s}=0\implies s=2

    According to the equation, s=-2 is also a solution, but we can ignore it because \{s_n\} is bounded below by 0.
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