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Math Help - convergence help

  1. #1
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    convergence help

    For each positive integer n, let

    {y_n}= 1 + 1/2 +...+1/n- the integral of (1/x) from 1 to n.

    Prove that the sequence {y_n} converges.
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  2. #2
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by friday616 View Post
    For each positive integer n, let

    {y_n}= 1 + 1/2 +...+1/n- the integral of (1/x) from 1 to n.

    Prove that the sequence {y_n} converges.
    Proof of monotone decreasing: y_{n-1}-y_n=\sum_{k=1}^{n-1}\frac{1}{k}-\sum_{k=1}^n\frac{1}{k}+\int_1^n\frac{1}{x}\,dx-\int_1^{n-1}\frac{1}{x}\,dx =\int_{n-1}^n\frac{1}{x}\,dx-\frac{1}{n}\geq\int_{n-1}^n\frac{1}{n}\,dx-\frac{1}{n}=0

    To prove that \{y_n\} is bounded below by 0, if P is the partition \{x_1=1,x_2=2,...,x_n=n\}, and f(x)=\frac{1}{x}, the upper sum U(P,f) of the integral of f(x) is \sum_{k=1}^{n-1}\frac{1}{x_k}(x_{k+1}-x_k)=\sum_{k=1}^{n-1}\frac{1}{k}. Since \sum_{k=1}^n\frac{1}{k}>U(P,f)>\int_1^n\frac{1}{x}  \,dx, it follows that y_n>0 for all n.

    Since \{y_n\} is a monotonically decreasing sequence bounded below, it converges.

    This is actually a famous limit and converges to the Euler-Mascheroni Constant, \gamma.

    \lim_{n\to\infty}\left(\sum_{k=1}^n\frac{1}{k}-\ln(n)\right)=\gamma\approx0.577

    You're welcome.
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