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Math Help - Sequences and Series Question

  1. #1
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    Sequences and Series Question

    Hi,

    I am new to this forum so please let me know if I am posting in the wrong place, and I will move my thread.

    I have the following question:
    Let the sequence (a_n) be given recursively by the formula:
    a_{n+1}=a_n+\frac{1}{n^2a_n} , where a_1=1


    i) Prove that a_n\ge1 for all natural numbers n
    ii) Prove that a_{n+1}\ge{a_n} for all natural numbers n
    iii) Prove that a_n\le1+\sum_{k=1}^{n-1} \frac{1}{k^2} for all n\ge2

    I have managed to do parts i) and ii), but am unable to do part iii)
    I think part i) is meant to help somehow.

    Help would be appreciated,

    Thanks
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  2. #2
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    Opalg's Avatar
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    Re: Sequences and Series Question

    Hint for (iii): induction (using (i), as you guessed).
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  3. #3
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    Re: Sequences and Series Question

    Quote Originally Posted by Opalg View Post
    Hint for (iii): induction (using (i), as you guessed).
    Is this correct?

    Proof by induction
    Prove for n=2
    a_2=a_{1+1}=a_1+\frac{1}{n^2a_1}=1+1=2
    1+\sum_{k=1}^1 \frac{1}{k^2}=1+1=2
    Case holds for n=2

    Assume case holds for n=q, i.e.
    a_q\le1+\sum_{k=1}^{q-1} \frac{1}{k^2}

    Prove for n=q+1
    a_{q+1}=a_q+\frac{1}{n^2a_q}
    a_{q+1}\le 1+\sum_{k=1}^{q-1} \frac{1}{k^2}+\frac{1}{n^2a_q}
    a_{q+1}\le a_{q+1}+\sum_{k=1}^{q-1} \frac{1}{k^2}+\frac{1}{n^2}
    0\le \sum_{k=1}^{q-1} \frac{1}{k^2}+\frac{1}{n^2}

    which is clearly true.
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  4. #4
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    Re: Sequences and Series Question

    Quote Originally Posted by dwally89 View Post
    Is this correct?

    Proof by induction
    Prove for n=2
    a_2=a_{1+1}=a_1+\frac{1}{n^2a_1}=1+1=2
    1+\sum_{k=1}^1 \frac{1}{k^2}=1+1=2
    Case holds for n=2

    Assume case holds for n=q, i.e.
    a_q\le1+\sum_{k=1}^{q-1} \frac{1}{k^2}

    Prove for n=q+1
    a_{q+1}=a_q+\frac{1}{n^2a_q}
    a_{q+1}\le 1+\sum_{k=1}^{q-1} \frac{1}{k^2}+\frac{1}{n^2a_q}
    a_{q+1}\le a_{q+1}+\sum_{k=1}^{q-1} \frac{1}{k^2}+\frac{1}{n^2}
    0\le \sum_{k=1}^{q-1} \frac{1}{k^2}+\frac{1}{n^2}

    which is clearly true.
    Anyone?
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  5. #5
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    Re: Sequences and Series Question

    Quote Originally Posted by dwally89 View Post
    Is this correct?

    Proof by induction
    Prove for n=2
    a_2=a_{1+1}=a_1+\frac{1}{n^2a_1}=1+1=2
    1+\sum_{k=1}^1 \frac{1}{k^2}=1+1=2
    Case holds for n=2

    Assume case holds for n=q, i.e.
    a_q\le1+\sum_{k=1}^{q-1} \frac{1}{k^2}

    Prove for n=q+1
    a_{q+1}=a_q+\frac{1}{n^2a_q}

    ******************************************

    What you should be doing from here is showing that "if"

    a_q\le 1+\sum_{k=1}^{q-1}\frac{1}{k^2}

    then

    a_{q+1}\le 1+\sum_{k=1}^{q}\frac{1}{k^2}

    ******************************************

    a_{q+1}\le 1+\sum_{k=1}^{q-1} \frac{1}{k^2}+\frac{1}{n^2a_q}
    a_{q+1}\le a_{q+1}+\sum_{k=1}^{q-1} \frac{1}{k^2}+\frac{1}{n^2}
    0\le \sum_{k=1}^{q-1} \frac{1}{k^2}+\frac{1}{n^2}

    which is clearly true.
    P(q);\;\;\;\;a_q\le 1+\sum_{k=1}^{q-1}\frac{1}{k^2}


    P(q+1);\;\;\;\;a_{q+1}\le 1+\sum_{k=1}^{q}\frac{1}{k^2}

    Show that "if" P(q) is true, "then" P(q+1) must be true.
    Write P(q+1) in terms of P(q).

    a_{q+1}=a_q+\frac{1}{q^2a_q}

    a_q+\frac{1}{q^2a_q}\le\ 1+\sum_{k=1}^{q}\frac{1}{k^2}\;\;?

    a_q+\frac{1}{q^2a_q}\le\ 1+\sum_{k=1}^{q-1}\frac{1}{k^2}+\frac{1}{q^2}\;\;?

    If P(q) is true, then

    a_q+\frac{1}{q^2a_q}\le\ a_q+\frac{1}{q^2}\;\;?

    Now, you may use the result to part (i) to justify the truth of this.
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