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Math Help - Infinite Series

  1. #1
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    Infinite Series

    Hi guys!

    I have the infinite series:

    \sum_{n=1}^\infty \frac{1}{\sqrt{n^2+3}}

    I know that I can use the integral test which involves using trig substitution. However, I know that there must be a p-series that I can compare it to. By looking at it, I know that it diverges. When I try to use \sum_{n=1}^\infty \frac{1}{n} which is a divergent p-series, and use the limit comparison test or direct comparison test, it doesn't work. I also tried using \sum_{n=1}^\infty \frac{1}{\sqrt{n}} but that does not work either. Does anyone have any ideas for a test series that I could compare my series to?

    PS. I don't need it worked out, just an idea of a test series

    Thanks in advance,

    Liz.
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  2. #2
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    Quote Originally Posted by elizsimca View Post
    Hi guys!

    I have the infinite series:

    \sum_{n=1}^\infty \frac{1}{\sqrt{n^2+3}}

    I know that I can use the integral test which involves using trig substitution. However, I know that there must be a p-series that I can compare it to. By looking at it, I know that it diverges. When I try to use \sum_{n=1}^\infty \frac{1}{n} which is a divergent p-series, and use the limit comparison test or direct comparison test, it doesn't work. I also tried using \sum_{n=1}^\infty \frac{1}{\sqrt{n}} but that does not work either. Does anyone have any ideas for a test series that I could compare my series to?

    PS. I don't need it worked out, just an idea of a test series

    Thanks in advance,

    Liz.
    For n \geq 1, \frac{1}{\sqrt{n^2 + 3}} > \frac{1}{\sqrt{n^2 + 3n^2}} = \frac{1}{2n}.
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  3. #3
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by elizsimca View Post
    Hi guys!

    I have the infinite series:

    \sum_{n=1}^\infty \frac{1}{\sqrt{n^2+3}}

    I know that I can use the integral test which involves using trig substitution. However, I know that there must be a p-series that I can compare it to. By looking at it, I know that it diverges. When I try to use \sum_{n=1}^\infty \frac{1}{n} which is a divergent p-series, and use the limit comparison test or direct comparison test, it doesn't work. I also tried using \sum_{n=1}^\infty \frac{1}{\sqrt{n}} but that does not work either. Does anyone have any ideas for a test series that I could compare my series to?

    PS. I don't need it worked out, just an idea of a test series

    Thanks in advance,

    Liz.
    also try \lim_{n\to\infty}\frac{\frac{1}{\sqrt{n^2+3}}}{\fr  ac{1}{n}}

    and since \frac{1}{\sqrt{n^2+3}}\sim\frac{1}{n}

    \lim_{n\to\infty}\frac{\frac{1}{\sqrt{n^2+3}}}{\fr  ac{1}{n}}=1

    and we know that \frac{1}{n} diverges by the integral test...so therefore since the limit converges to a value both share converence or divergence...so both are divergent
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by mr fantastic View Post
    For n \geq 1, \frac{1}{\sqrt{n^2 + 3}} > \frac{1}{\sqrt{n^2 + 3n^2}} = \frac{1}{2n}.
    Another one would be \frac{1}{\sqrt{n^2+3}}<\frac{1}{n}

    Same conclusion
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  5. #5
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    Quote Originally Posted by Mathstud28 View Post
    Another one would be \frac{1}{\sqrt{n^2+3}}<\frac{1}{n}

    Same conclusion
    I don't think we can't use this one. \frac{1}{n} diverges, therefore we would have to prove that for some n>N \frac{1}{\sqrt{n^2+3}}\geq \frac{1}{n} to prove divergence of our unknown series, which is not true. I think..right?

    But the post above that you left, evaluating the limit and discovering it is a finite value proves that they have the same behavior..I like it!
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  6. #6
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    Quote Originally Posted by mr fantastic View Post
    For n \geq 1, \frac{1}{\sqrt{n^2 + 3}} > \frac{1}{\sqrt{n^2 + 3n^2}} = \frac{1}{2n}.
    You do realise that, by the comparison test, this means your series is divergent, I hope .....
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  7. #7
    Super Member angel.white's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    Another one would be \frac{1}{\sqrt{n^2+3}}<\frac{1}{n}

    Same conclusion
    I may be confused, but if \frac{1}{\sqrt{n^2+3}}<\frac{1}{n} then showing that 1/n diverges doesn't prove anything. I always thought of it as the series diverging must be "pushing" the series it is being compared to (if it diverges, and "holding" or "bouding" the series it is being compared to if it converges). So you would need to use Mr. F's method, or one similar to it.
    Quote Originally Posted by Mathstud28 View Post
    also try \lim_{n\to\infty}\frac{\frac{1}{\sqrt{n^2+3}}}{\fr  ac{1}{n}}

    and since \frac{1}{\sqrt{n^2+3}}\sim\frac{1}{n}

    \lim_{n\to\infty}\frac{\frac{1}{\sqrt{n^2+3}}}{\fr  ac{1}{n}}=1

    and we know that \frac{1}{n} diverges by the integral test...so therefore since the limit converges to a value both share converence or divergence...so both are divergent
    A good way of doing it, but I think a more rigorous way of showing this would be:
    \lim_{n\to\infty}\frac{\frac{1}{\sqrt{n^2+3}}}{\fr  ac{1}{n}}~~~~=~~~~\lim_{n\to\infty} \frac{n}{\sqrt{n^2+3}}~~~~=~~~~\lim_{n\to\infty} \frac{1}{\sqrt{1+\frac 3{n^2}}}~~~~=~~~~1
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  8. #8
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    Quote Originally Posted by Mathstud28 View Post
    Another one would be \frac{1}{\sqrt{n^2+3}}<\frac{1}{n}

    Same conclusion
    Quote Originally Posted by angel.white View Post
    I may be confused, but if \frac{1}{\sqrt{n^2+3}}<\frac{1}{n} then showing that 1/n diverges doesn't prove anything.
    [snip]
    You're not totally confused. Au contraire, you're right on the money.

    A comparison based on the inequality \frac{1}{\sqrt{n^2+3}}<\frac{1}{n} does not work. If you suspect divergence, you require an inequality such that \frac{1}{\sqrt{n^2+3}} {\color{red}>} v_n where \sum v_n is divergent.


    And for completeness, if you suspect convergence, you require an inequality such that \frac{1}{\sqrt{n^2+3}} {\color{red}<} v_n where \sum v_n converges..


    If you cannot find an inequality that confirms your suspicion, then you're suspicion is most likely wrong.
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  9. #9
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    Quote Originally Posted by mr fantastic View Post
    You do realise that, by the comparison test, this means your series is divergent, I hope .....
    Yeah, I said that in the original question
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  10. #10
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    Quote Originally Posted by mr fantastic View Post
    You're not totally confused. Au contraire, you're right on the money.

    A comparison based on the inequality \frac{1}{\sqrt{n^2+3}}<\frac{1}{n} does not work. If you suspect divergence, you require an inequality such that \frac{1}{\sqrt{n^2+3}} {\color{red}>} v_n where \sum v_n is divergent.


    And for completeness, if you suspect convergence, you require an inequality such that \frac{1}{\sqrt{n^2+3}} {\color{red}<} v_n where \sum v_n converges..


    If you cannot find an inequality that confirms your suspicion, then you're suspicion is most likely wrong.
    O sorry..it was late and I was tired...yes you are correct
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