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Math Help - Series problems

  1. #1
    Newbie usvn's Avatar
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    Series problems

    I need some help with these series problems, i do not know what test to use on them. Help appreciated thanks (also i do not know how to type a summation symbol).



    Summation n=0 to infinity 13arctan(n)

    summation n=1 to infinity sin(43n) / 1+n^2
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  2. #2
    o_O
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    #1: Use the divergence test. If \lim_{n \to \infty} a_n \neq 0, then the \sum a_n diverges.

    #2: Use the fact that if \sum |a_n| converges, then so does \sum a_n. Then make use of the comparison test by using |\sin x | \leq 1
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  3. #3
    Newbie usvn's Avatar
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    <br /> <br />
\lim_{n \to \infty} 13arctan(n)<br />
does not go to 0 so divergent(wow simple).

    I know this probebly isent right but using the direct comparison test the best i could come up with is
    <br />
\sin (43n)/1+n^2  \le 2/n^2<br />

    less then a p series that is convergent meaning convergent?
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  4. #4
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    #1:

    #2: You can't use a direct comparison because one of the conditions of the comparison test is that the terms of your sequence has to be positive for all integers after some nth term.

    This is why we consider the sequence |a_n| where all terms are obviously positive and is why the theorem I gave you is useful.

    So: \sum \left| \frac{\sin (43n)}{1+n^2} \right| = \sum \frac{|\sin (43n)|}{1+n^2} {\color{red} \ \leq \ } \sum \frac{1}{1+n^2} < \cdots

    Can you finish?
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  5. #5
    Newbie usvn's Avatar
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    I would then show it is less then a convergent p series yes, but is that really less or equal because it has a numerator which would make it larger then the one without one... wouldn't it?
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  6. #6
    o_O
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    Think about it this way: |\sin (43n)| \leq 1

    Multiply both sides by \frac{1}{1+n^2} > 0 to get: \frac{|\sin (43n)|}{1+n^2} \leq \frac{1}{1+n^2}
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  7. #7
    Newbie usvn's Avatar
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    Thanks, i realize sin(x) will never be greater then one so that would be possible.
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