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Math Help - Series and sequences

  1. #1
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    Exclamation Series and sequences

    I'm having problem with this question.

    Do I use the ratio test for this infinite series? if so, can you show a few steps to get me going, or if not, what should I use?

    \sum\limits_{n=1}^{\infty} \frac{2^n - n^5}{4^n+2n^2+3n}

    Thanks in advance
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  2. #2
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    Quote Originally Posted by tsal15 View Post
    I'm having problem with this question.

    Do I use the ratio test for this infinite series? if so, can you show a few steps to get me going, or if not, what should I use?

    \sum\limits_{n=1}^{\infty} \frac{2^n - n^5}{4^n+2n^2+3n}

    Thanks in advance
    You have \sum\limits_{n=1}^{\infty} \frac{2^n}{4^n+2n^2+3n} - \sum\limits_{n=1}^{\infty} \frac{n^5}{4^n+2n^2+3n}.

    Now note:

    1. \frac{2^n}{4^n+2n^2+3n} < \frac{2^n}{4^n} = \left(\frac{1}{2}\right)^n.

    So the first sum converges by the comparison test.


    2. \frac{n^5}{4^n+2n^2+3n} < \frac{n^5}{4^n} and \sum\limits_{n=1}^{\infty} \frac{n^5}{4^n} converges by the ratio test.

    So the second sum converges.
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  3. #3
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    Quote Originally Posted by mr fantastic View Post
    You have \sum\limits_{n=1}^{\infty} \frac{2^n}{4^n+2n^2+3n} - \sum\limits_{n=1}^{\infty} \frac{n^5}{4^n+2n^2+3n}.

    Now note:

    1. \frac{2^n}{4^n+2n^2+3n} < \frac{2^n}{4^n} = \left(\frac{1}{2}\right)^n.

    So the first sum converges by the comparison test.


    2. \frac{n^5}{4^n+2n^2+3n} < \frac{n^5}{4^n} and \sum\limits_{n=1}^{\infty} \frac{n^5}{4^n} converges by the ratio test.

    So the second sum converges.

    Thank you for the quick reply Mr. Fantastic. You truly are fantastic .
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  4. #4
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    Thumbs up

    Now note:

    1. \frac{2^n}{4^n+2n^2+3n} < \frac{2^n}{4^n} = \left(\frac{1}{2}\right)^n.

    So the first sum converges by the comparison test.
    Oh Mr. Fantastic I should've asked you this earlier...I was just soo excited someone actually replied to my post... you see i haven't had much help for a while on mathhelp...but any who. how did you know to compare that sum with \frac{2^n}{4^n}?

    2. \frac{n^5}{4^n+2n^2+3n} < \frac{n^5}{4^n} and \sum\limits_{n=1}^{\infty} \frac{n^5}{4^n} converges by the ratio test.

    So the second sum converges.
    I believe you've used the comparison test, again, after finding out \frac{n^5}{4^n} converges by the ratio test? Similarly, how did you know to use \frac{n^5}{4^n}

    Thank you Mr. Fantastic
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  5. #5
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    Quote Originally Posted by tsal15 View Post
    Oh Mr. Fantastic I should've asked you this earlier...I was just soo excited someone actually replied to my post... you see i haven't had much help for a while on mathhelp...but any who. how did you know to compare that sum with \frac{2^n}{4^n}?

    I believe you've used the comparison test, again, after finding out \frac{n^5}{4^n} converges by the ratio test? Similarly, how did you know to use \frac{n^5}{4^n} Mr F says: Yes, I did.

    Thank you Mr. Fantastic
    In both cases it's a matter of guessing whether the series is convergent or divergent and then trying to construct an appropriate series to use in the comparison. The key is a lot of practice to develop experience - after a while you get to 'see' (sometimes) what the series will be that you need to use.
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  6. #6
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    Quote Originally Posted by mr fantastic View Post
    In both cases it's a matter of guessing whether the series is convergent or divergent and then trying to construct an appropriate series to use in the comparison. The key is a lot of practice to develop experience - after a while you get to 'see' (sometimes) what the series will be that you need to use.
    hmmm, guessing i can get used to that...hehehe. but yes i do agree with, practice will help.

    thanks Mr. Fantastic
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