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Math Help - Sequence

  1. #1
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    Sequence

    Hi everybody,

    U_n=1+\frac{1}{2^3}+\frac{1}{3^3}+\frac{1}{4^3}+..  .+\frac{1}{n^3}

    I must show that (\forall n \in \mathbb{N}^*) U_n\le 2-\frac{1}{n}

    I used recurrence but i didn't find anything.
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  2. #2
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    Quote Originally Posted by bhitroofen01 View Post
    Hi everybody,

    U_n=1+\frac{1}{2^3}+\frac{1}{3^3}+\frac{1}{4^3}+..  .+\frac{1}{n^3}

    I must show that (\forall n \in \mathbb{N}^*) U_n\le 2-\frac{1}{n}

    I used recurrence but i didn't find anything.
    Use induction:

    The base case n=1 is true.

    Assume true for n=k, then

    U_{k+1}=U_k+\frac{1}{(k+1)^3}\le 1-\frac{1}{k}+\frac{1}{(k+1)^3}

    and with a bit of work this can be manipulated to show that:

    U_{k+1}\le 2 - \frac{1}{k+1}

    which allows one to claim proof bt induction

    CB
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  3. #3
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    And how to show that (U_n) is convergent

    I know just that \lim_{n\to +\infty} U_n \le 2 \ \ ????
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  4. #4
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    The integral test can be used to show that \sum_{n=1}^\infty \frac{1}{n^r} converges for r> 1.
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  5. #5
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    Quote Originally Posted by bhitroofen01 View Post
    And how to show that (U_n) is convergent

    I know just that \lim_{n\to +\infty} U_n \le 2 \ \ ????
    U_n is an increasing bounded sequence and hence converges, and so the series converges.

    CB
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