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Thread: Convergence

  1. #1
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    Convergence

    Show $\displaystyle \left(\frac{1}{n^k}\right)_{n\in\mathbb{N}$ is convergent iff. $\displaystyle k\geq 0$, and that $\displaystyle \left(\frac{1}{n^k}\right)\rightarrow 0, \ \forall k>0$

    (i) assume $\displaystyle \left(\frac{1}{n^k}\right)$ is convergent.
    Because it is convergent, there is some value alpha such that:
    $\displaystyle \left(\frac{1}{n^k}\right)\rightarrow\alpha$
    $\displaystyle \Rightarrow\left |\frac{1}{n^k}-\alpha\right |<\epsilon$

    I am not sure if I am on the right track, and if I am, what should I do next?
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  2. #2
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    For the forward implication I would break the proof into two parts.

    First show that

    $\displaystyle k > 0$

    I assume that you mean the above if k=0 the sequence is constant and equal to 1.

    So suppose not( assume k < 0) and use what you have above to get a contradiction. Hint the sequence will be unbounded.

    Then after you know that k > 0 you can make a direct argument to show that it converges to zero.

    The reverse <= implication is not bad at all.
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  3. #3
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    Instead of doing "k non-negative implies convergent" and "convergent implies k non-negative", it'll probably be easier to do "k non-negative implies convergent" and "k negative implies divergent".

    There are probably nicer ways to do it, but for k non-negative, I'd choose m such that 1/m<k. Then $\displaystyle \sqrt[m]{n}<n^k$, and since you know $\displaystyle \sqrt[m]n\to\infty$ as $\displaystyle n\to\infty$, so must $\displaystyle n^k$, and so $\displaystyle \frac1{n^k}\to0$. The case where k is negative follows immediately.
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  4. #4
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    (a) by contradiction:
    $\displaystyle \left(\frac{1}{n^k}\right)$ is convergent and k < 0.
    Since k < 0, we can right $\displaystyle k=-r, \ r\in\mathbb{R}, \ r>0$
    $\displaystyle \Rightarrow\left(\frac{1}{n^k}\right)=\left(\frac{ 1}{n^{-r}}\right)=(n^r)$
    $\displaystyle \Rightarrow\text{as} \ n\rightarrow\infty \ (n^r)\rightarrow\infty$
    Thus, we have reached contradiction and the sequence is convergent for $\displaystyle k\geq 0$
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  5. #5
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    (b) $\displaystyle \left(\frac{1}{n^k}\right)\rightarrow 0, \ \forall k>0$
    $\displaystyle \left(\frac{1}{n^k}\right) \ \text{and as} \ n\to\infty, \ \left(\frac{1}{\infty^k}\right)=\left(\frac{1}{ \infty }\right)\to 0$

    Good?
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  6. #6
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    Quote Originally Posted by dwsmith View Post
    (a) by contradiction:
    $\displaystyle \left(\frac{1}{n^k}\right)$ is convergent and k < 0.
    Since k < 0, we can right $\displaystyle k=-r, \ r\in\mathbb{R}, \ r>0$
    $\displaystyle \Rightarrow\left(\frac{1}{n^k}\right)=\left(\frac{ 1}{n^{-r}}\right)=(n^r)$
    $\displaystyle \Rightarrow\text{as} \ n\rightarrow\infty \ (n^r)\rightarrow\infty$
    Thus, we have reached contradiction and the sequence is convergent for $\displaystyle k\geq 0$
    Since this is for an analysis class I would be a bit more explicit.

    Since we know that

    $\displaystyle \left\{\frac{1}{n^k} \right\} \to 0$

    This implies that for every epsilon greater than 0 there exists an N such that ... so set

    $\displaystyle \epsilon=\frac{1}{2} \implies \exists N \in \mathbb{N} \text{ such that }\left|\frac{1}{n^k}-\alpha \right| < \frac{1}{2} \implies -\frac{1}{2}+\alpha < \frac{1}{n^k} < \frac{1}{2}+\alpha$

    But this implies that the sequence is bounded. But since k < 0 we can make

    $\displaystyle \left\{\frac{1}{n^k} \right\}$

    as large as we wish.

    Let M > 0 be any real number

    Then Pick

    $\displaystyle N'=\left( \frac{1}{M}\right)^{\frac{1}{k}}$

    Then for any n > n'

    $\displaystyle \left\{\frac{1}{n^k} \right\} > M$

    so we have the desired contradiction.
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  7. #7
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    Quote Originally Posted by TheEmptySet View Post
    Let M > 0 be any real number

    Then Pick

    $\displaystyle N'=\left( \frac{1}{M}\right)^{\frac{1}{k}}$

    Then for any n > n'

    $\displaystyle \left\{\frac{1}{n^k} \right\} > M$

    so we have the desired contradiction.
    I don't quite understand what is going on here.
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  8. #8
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    Quote Originally Posted by dwsmith View Post
    I don't quite understand what is going on here.
    Lets look at a concreate example.

    Let

    $\displaystyle k=-\frac{1}{2}$

    and set

    $\displaystyle M=1,000,000=10^6$

    Now notice if I set

    $\displaystyle N'=\left( \frac{1}{10^6} \right)^{ \frac{1}{-\frac{1}{2}} $ $\displaystyle =10^{12}$

    Now for any

    $\displaystyle n > N'=10^{12}$

    $\displaystyle a_{10^{12}}=\frac{1}{(10^{12})^k}=10^{-12k}=10^{6} $

    if $\displaystyle k=-\frac{1}{2}$

    This shows that if k is negative the sequced can be made as large as we want. There is nothing special about one million the above is true for any positive M so the sequence diverges to infintity.
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