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Thread: 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 $\displaystyle (a_n)$ be given recursively by the formula:
    $\displaystyle a_{n+1}=a_n+\frac{1}{n^2a_n}$ , where $\displaystyle a_1=1$


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

    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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    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 $\displaystyle n=2$
    $\displaystyle a_2=a_{1+1}=a_1+\frac{1}{n^2a_1}=1+1=2$
    $\displaystyle 1+\sum_{k=1}^1 \frac{1}{k^2}=1+1=2$
    Case holds for $\displaystyle n=2$

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

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

    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 $\displaystyle n=2$
    $\displaystyle a_2=a_{1+1}=a_1+\frac{1}{n^2a_1}=1+1=2$
    $\displaystyle 1+\sum_{k=1}^1 \frac{1}{k^2}=1+1=2$
    Case holds for $\displaystyle n=2$

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

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

    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 $\displaystyle n=2$
    $\displaystyle a_2=a_{1+1}=a_1+\frac{1}{n^2a_1}=1+1=2$
    $\displaystyle 1+\sum_{k=1}^1 \frac{1}{k^2}=1+1=2$
    Case holds for $\displaystyle n=2$

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

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

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

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

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

    then

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

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

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

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


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

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

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

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

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

    If P(q) is true, then

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

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