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Thread: Why can't this be proved through comparison/limit comparison test?

  1. #1
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    Why can't this be proved through comparison/limit comparison test?

    My books says if the following series can not be proved convergent or divergent (even if they can be shown through other tests) through the limit comparison or direct comparison test, then write "NA". If they can be proved, then write CONV or DIV.

    The series is as follows:
     a_n= \sum_{n=1}^{\infty} \frac{cos(n)\sqrt{n}}{2n+9}

    The books says that that series can't be proved DIV or CONV through the comparison tests.

    I have a way, but can't see whats wrong with it. Here is my solution:

     \frac{cos(n) \sqrt{n}}{2n+9} < \frac{\sqrt{n}}{2n+9} let the new series be b_n (because cos(n) will always be negative and in between -1 and 1.... meaning the numerator will always be a lower or equal value than the new series without the cosine. Then with  \frac{\sqrt{n}}{2n+9} \Rightarrow \frac{1}{n^\frac{3}{2} + 9} (let this new series be c_n) and this is convergent because as it is a p-series (if you get rid of the 9 which will be irrelevant as  n \rightarrow \infty ... and so through the direct comparison test, the original series should be convergent because:  a_n < b_n < c_n and c_n is convergent. Where am I going wrong?
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    Re: Why can't this be proved through comparison/limit comparison test?

    Hi,
    The comparison test or limit comparison test is applicable only to series of non-negative terms. With a slight modification of your argument, you have proved the series is absolutely convergent and hence convergent. But comparison test is not directly applicable.
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    Re: Why can't this be proved through comparison/limit comparison test?

    hmm, can you elaborate on the slight modification? What can I change to make my proof work?
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    Re: Why can't this be proved through comparison/limit comparison test?

    Just say $|\cos(n)|\leq 1$; all other parts of the terms are non-negative.
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    Re: Why can't this be proved through comparison/limit comparison test?

    Are we to understand that the series in question is the following? $$\sum_{n = 1}^\infty a_n \qquad \text{where} \, a_n = {\sqrt{n}\cos n \over n+9}$$

    In terms of absolute convergence, we can say that $\sum a_n$ converges if $\sum |a_n|$ converges. But we have a problem in proving that.
    $$0 \lt \left|{\sqrt{n}\cos n \over n+9}\right| \lt {\sqrt{n} \over n+9} \lt {\sqrt{n} \over n} = {1 \over \sqrt{n}} = b_n$$
    But the series $\sum b_n$ is not convergent anyway.

    We might like to use a comparison with a series like $\sum {1 \over n}$ to prove divergence, but then the fact that some of the terms of the original sequence are negative and some are probably much smaller than ${1 \over n}$ scuppers that plan.

    The ${1 \over n^{3 \over 2} + 9}$ in the original post is inaccurate.

    Wolfram Alpha suggests that the series might converge. And, being an approximately alternating series with terms tending to zero, it is likely to converge.
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    Re: Why can't this be proved through comparison/limit comparison test?

    Quote Originally Posted by Archie View Post
    Are we to understand that the series in question is the following? $$\sum_{n = 1}^\infty a_n \qquad \text{where} \, a_n = {\sqrt{n}\cos n \over n+9}$$

    In terms of absolute convergence, we can say that $\sum a_n$ converges if $\sum |a_n|$ converges. But we have a problem in proving that.
    $$0 \lt \left|{\sqrt{n}\cos n \over n+9}\right| \lt {\sqrt{n} \over n+9} \lt {\sqrt{n} \over n} = {1 \over \sqrt{n}} = b_n$$
    But the series $\sum b_n$ is not convergent anyway.

    We might like to use a comparison with a series like $\sum {1 \over n}$ to prove divergence, but then the fact that some of the terms of the original sequence are negative and some are probably much smaller than ${1 \over n}$ scuppers that plan.

    The ${1 \over n^{3 \over 2} + 9}$ in the original post is inaccurate.

    Wolfram Alpha suggests that the series might converge. And, being an approximately alternating series with terms tending to zero, it is likely to converge.
    ahh yes, how stupid of me.
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