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Math Help - Summon the Heroes-2

  1. #1
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    Problem 2

    Assume \sum_{n=1}^{\infty}a_{n} converges

    where a_{n}>0 and a_{n} \geq a_{n+1} for \forall n \in \mathbb{N}^+


    show that: \lim_{n \rightarrow \infty}na_{n}=0
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  2. #2
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    Quote Originally Posted by Xingyuan View Post
    Assume \sum_{n=1}^{\infty}a_{n} converges

    where a_{n}>0 and a_{n} \geq a_{n+1} for \forall n \in \mathbb{N}^+


    show that: \lim_{n \rightarrow \infty}na_{n}=0
    Suppose that \sum_{n=1}^{\infty}a_{n} = S. Given \varepsilon>0, choose N so that \sum_{r=1}^{N}a_r > S-\tfrac12\varepsilon. Notice that if n>2N it follows that n-N>\tfrac12n.

    If n>2N then \tfrac12\varepsilon > \sum_{r=N+1}^{n}a_r >(n-N)a_n (since the terms a_r are decreasing), and so \tfrac12\varepsilon > \tfrac12na_n. Thus n>2N\;\Longrightarrow\;na_n<\varepsilon. Therefore na_n\to0 as n\to\infty.
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  3. #3
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    I've found a nice proof without \varepsilon's...

    Note that the sequence of general term u_n=\sum_{k=1}^n (a_k-a_n), n\geq 1, is nondecreasing. This is because the terms of the sum are nonnegative ( a_n\leq a_k for k\leq n), and each of them (i.e. a_k-a_n) increases with n.

    Plus, u_n\leq \sum_{k=1}^\infty a_k<\infty, hence (u_n)_{n\geq 1} converges to a finite limit.

    However, u_n=\sum_{k=1}^n a_k - na_n and the sum converges to a finite limit as n\to\infty, hence na_n must converge to a finite limit \alpha as well.

    This limit has to be 0, for otherwise a_k\sim \frac{\alpha}{k} is the general term of a divergent series. Thus, n a_n\to 0.
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  4. #4
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    Ingenious Work !!
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