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Math Help - Testing series for convergence

  1. #1
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    Testing series for convergence

    Asked to test the following for convergence using the comparason test as much as possible.

    1. 1/(n^p-n^q) 0<q<p I thin kthe limit in this one does not exist but cant say why.

    2. sqrt(N^2+1) - n

    3. (n^(1/n) -1) I think this converges to a limit of -1 but cant show why


    Please help I am totally stuck.

    Regards
    Niall101
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  2. #2
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    Your notation is not so clear.
    Do you mean for example number 2 is showing the convergence/divergence of \sum \sqrt{n^2+1}-n?

    \sqrt{n^2+1}-n=\frac{1}{\sqrt{n^2+1}+n}>\frac{1}{(n+1)+n}>\frac  {1}{2}\frac{1}{n+1}
    So the series is divergent.
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  3. #3
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    Quote Originally Posted by Niall101 View Post
    1. 1/(n^p-n^q) 0<q<p I thin kthe limit in this one does not exist but cant say why.

     <br />
\frac{1}{n^p-n^q}=\frac{1}{n^q(n^{p-q}-1)}<br />

    but p-q<0, so:

     <br />
\lim_{n \to \infty} \frac{1}{n^{p-q}-1}=-1<br />

    and as q>0 :

     <br />
\lim_{n \to \infty} \frac{1}{n^q}=0<br />

    So:

     <br />
\lim_{n \to \infty}\frac{1}{n^p-n^q}=0<br />

    CB
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  4. #4
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    Quote Originally Posted by Niall101 View Post
    2. sqrt(N^2+1) - n

     <br />
\lim_{n \to \infty} \sqrt{n^2+1} - n = \lim_{n \to \infty} n(1+n^{-2})^{1/2}-n<br />

    Now expand the square root as a power series to see what the limit is

    CB
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  5. #5
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    The tittle seems that we are talking about the convergence/divergence of series here.
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  6. #6
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    Quote Originally Posted by Niall101 View Post
    3. (n^(1/n) -1) I think this converges to a limit of -1 but cant show why
    Not surprised as it goes to 0.

    The -1 makes this more complicated than it need be so lose it:

    \lim_{n \to \infty} (n^{1/n} -1)=[\lim_{n \to \infty} n^{1/n}] -1

    Now you could find the limit on the right by looking at the limit of its log.

    RonL
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  7. #7
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    Quote Originally Posted by watchmath View Post
    The tittle seems that we are talking about the convergence/divergence of series here.
    If one finds the limits and they are finite then they converge. To find all the limits may not be what was intended but it is convincing and proves what was asked if not by the means intended.

    It also simplifies the process of testing if you know the answer (that is if they do or do not converge)

    Also, at a number of points in my posts one can just use tests to check for convergence rather than proceed to actualy find the limits.

    CB
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  8. #8
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    Quote Originally Posted by CaptainBlack View Post
    If one finds the limits and they are finite then they converge. To find all the limits may not be what was intended but it is convincing and proves what was asked if not by the means intended.

    It also simplifies the process of testing if you know the answer (that is if they do or do not converge)

    Also, at a number of points in my posts one can just use tests to check for convergence rather than proceed to actualy find the limits.

    CB
    I am not sure if you understand what I meant and I understand what you mean.
    The title of the thread is about the convergence of series (NOT sequence) and also he/she talked about comparison test.
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