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Math Help - Summon the Heroes-3

  1. #1
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    Thumbs up Summon the Heroes-3

    Problem 3

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

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

    show that: \sum_{n=1}^{\infty}a_{n}^2 converges
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  2. #2
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    Quote Originally Posted by Xingyuan View Post
    Problem 3

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

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

    show that: \sum_{n=1}^{\infty}a_{n}^2 converges

    Excuse the doubt, but all these "summon the heroes" exercises look suspiciously similar to rather standard exercises in calculus I and II...
    Are you sure these aren't "summon the guys that are going to do my homework for me" exercises? I've seen questions asked by you some few weeks ago and they did look like homework and not summonheroeds...

    Tonio
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  3. #3
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    Quote Originally Posted by Xingyuan View Post
    Problem 3

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

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

    show that: \sum_{n=1}^{\infty}a_{n}^2 converges
    Please show your attempt at solving this question and say where you get stuck.
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  4. #4
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    Lightbulb

    I only want to make some fun in solving math problems(like playing game)
    do not finish my homework.....
    actually ,I am not a student now...
    Thanks very much
    My idea is:

    As \sum_{n=1}^{\infty}\frac{a_{n}}{\sqrt{n}} converges

    so I think for massive(not all) n, \frac{a_{n}}{\sqrt{n}}\leq \frac{1}{n^{1+\epsilon}}

    then will give the conclusion

    But now I must to prove some a_{k} satisfy the condition: \frac{a_{k}}{\sqrt{k}}\geq \frac{1}{k^{1+\epsilon}}, do not
    make any essential contribution to the summation \sum_{n=1}^{\infty}{a_{n}}^2
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  5. #5
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    Note that \left(\frac{a_n}{\sqrt{n}}\right)_n is decreasing. Therefore, the problem is equivalent to:

    If \sum_n a_n converges, where (a_n)_n decreases, then \sum_n n a_n^2 converges.

    (Applying this to \frac{a_n}{\sqrt{n}} gives the initial statement)

    Using the property from your previous "challenge", the proof is then straightforward.

    (You can also directly use this property without rewriting the statement: n\frac{a_n}{\sqrt{n}}\to_n 0 hence \sqrt{n} a_n\leq M and a_n^2\leq a_n \frac{M}{\sqrt{n}}, qed)
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