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Math Help - convergence

  1. #1
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    convergence

    Use an appropriate test to determine the absolutely convergence, conditionally convergence or divergence of the series. Identify the test used.

    \sum_{n=2}^{\infty} \frac{cos^2(n+1)}{2^n +n}

    This was on a past test and my teacher doesn't provide us with the answers so can someone help solve this for me
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  2. #2
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    This problem is easier than you think. The series doesn't alternate (it will always be positive - you can figure out why on your own). The function f(n) also aproaches zero, as n tends to infinity as the numerator will always be a number between 1 and 0 (there maybe some fluctuation - but you are approaching zero). You can use the comparison test, and notice that the similar function is 1/2^n (which is convergent).
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  3. #3
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    Quote Originally Posted by qzno View Post
    Use an appropriate test to determine the absolutely convergence, conditionally convergence or divergence of the series. Identify the test used.

    \sum_{n=2}^{\infty} \frac{cos^2(n+1)}{2^n +n}

    This was on a past test and my teacher doesn't provide us with the answers so can someone help solve this for me
    It's fairly long, but first, it's a positive termed sequence, so you don't need to test for absolute or conditional convergence. Just test for convergence.


    Also note that 0 \leq \cos^2(n + 1) \leq 1.

    This would mean that

    \frac{\cos^2(n + 1)}{2^n + 1} \leq \frac{1}{2^n + 1}.

    We can say that if the "larger" series converges, so must the "smaller" series. So use the comparison test.


    \sum_{n = 2}^\infty\frac{\cos^2(n + 1)}{2^n + 1} \leq \sum_{n = 2}^\infty\frac{1}{2^n + 1}.


    Now applying the ratio test...

    a_k = \frac{1}{2^k + 1}

    a_{k + 1} = \frac{1}{2^{k + 1} + 1}


    \frac{a_{k + 1}}{a_k} = \frac{2^k + 1}{2^{k + 1} + 1}.

     = \frac{1}{2} + \frac{1}{2\left(2^{k + 1} + 1\right)}.


    Now letting k \to \infty we see

    \lim_{k \to \infty}\frac{a_{k + 1}}{a_k} = \frac{1}{2}.


    Since this limit <1, the series must converge.


    And since our "larger" series converges, so must the "smaller" series.
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    Thank you for that but I'm just wondering if there's any other way to figure it out of do you have to use those specific methods
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    Since you've already been shown two correct methods, I'd say there are lots of methods that can be used.
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  6. #6
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    note that \frac{1}{2^{n}+1}<\frac{1}{2^{n}}, so direct comparison applies faster.
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