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Math Help - divergent or convergent but turns out undefined?

  1. #1
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    divergent or convergent but turns out undefined?

    I am trying to find divergence or convergence using Maple of;

    SUM from n = 1 to inf of the equation:

    (1+sin(n)/n)^2

    I run it in Maple as a ratio test but when I try to take the limit, I get that the solution is undefined?
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    Quote Originally Posted by orendacl View Post
    I am trying to find divergence or convergence using Maple of;

    SUM from n = 1 to inf of the equation:

    (1+sin(n)/n)^2

    I run it in Maple as a ratio test but when I try to take the limit, I get that the solution is undefined?
    You can't evaluate \lim_{n \to \infty}\frac{\sin{n}}{n}, so I don't believe that you can determine convergence or divergence using the ratio test...
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    Quote Originally Posted by orendacl View Post
    I am trying to find divergence or convergence using Maple of;

    SUM from n = 1 to inf of the equation:

    (1+sin(n)/n)^2

    I run it in Maple as a ratio test but when I try to take the limit, I get that the solution is undefined?
    In fact, by the simple divergence test...

    \lim_{n \to \infty}\left(1 + \frac{\sin{n}}{n}\right)^2 = 1. Since this does not \to 0, the series is divergent.
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    My apologies.
    I really need to figure out how to get the fancy equations in these postings.
    The actual question is:

    [(1+sin(n))/n]^2

    Not:

    (1+sin(n)/n)^2

    Now, that probably changes things!
    I agree, that in the last scenario where "1" was not part of the numerator...then it would be divergent but what if we use the new equation - again, my apologies for lack of expertise in posting the question correctly the first time.
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    Quote Originally Posted by Prove It View Post
    You can't evaluate \lim_{n \to \infty}\frac{\sin{n}}{n},
    Use the sandwich theorem!
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    Quote Originally Posted by orendacl View Post
    I really need to figure out how to get the fancy equations in these postings.
    Why not learn to post in symbols? You can use LaTeX tags.
    [tex]\left[\frac{1+\sin(x)}{n}\right]^2[/tex] gives  \left[\frac{1+\sin(x)}{n}\right]^2 .
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  7. #7
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    OK, here it goes:

    \left[\frac{1+\sin(n)}{n}\right]^2

    that is my equation:

    I get this as being undefined or that...there is no convergence or divergence for this problem?
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    Note that \left[1+\sin(n)\right]^2\le4.
    So \sum\limits_{n = 1}^\infty  {\left[ {\frac{{1 + \sin (n)}}<br />
{n}} \right]^2 }  \leqslant \sum\limits_{n = 1}^\infty  {\frac{4}<br />
{{n^2 }}}
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  9. #9
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    Quote Originally Posted by Prove It View Post
    You can't evaluate \lim_{n \to \infty}\frac{\sin{n}}{n}, so I don't believe that you can determine convergence or divergence using the ratio test...
    What? since sin(x) is always between -1 and 1 and n goes to infinity, \lim_{n\to \infty}\frac{sin(n)}{n}= 0
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    I agree...You "can" evaluate that limit, and it does = 0
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    I think this goes infinity
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    Quote Originally Posted by HallsofIvy View Post
    What? since sin(x) is always between -1 and 1 and n goes to infinity, \lim_{n\to \infty}\frac{sin(n)}{n}= 0
    Thanks, I assumed that since \lim_{n \to \infty}\sin{n} doesn't exist, due to its oscillatory nature, then neither would \lim_{n \to \infty}\frac{\sin{n}}{n}.

    What happens at the points where you end up with \frac{0}{\infty} though?
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    Quote Originally Posted by Prove It View Post
    Thanks, I assumed that since \lim_{n \to \infty}\sin{n} doesn't exist, due to its oscillatory nature, then neither would \lim_{n \to \infty}\frac{\sin{n}}{n}.

    What happens at the points where you end up with \frac{0}{\infty} though?
    That still 0- you have very very small numbers divided by very very large numbers- even closer to 0!
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  14. #14
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    Yes but,
    The terms of our series are strictly between 0 and 4/n^2.
    "both of those series" WILL converge so shouldn't we then just use the comparison test?
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    Quote Originally Posted by orendacl View Post
    Yes but,
    The terms of our series are strictly between 0 and 4/n^2.
    "both of those series" WILL converge so shouldn't we then just use the comparison test?
    Absolutely correct.
    The series converges by the basic comparison test.
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