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Check for the convergence or divergence of the following series

  1. Oct 4, 2013 #1
    1. The problem statement, all variables and given/known data
    Here are some series I'm completely stuck on.
    1.sqrt(n)*(1-cos(1/n))

    2. a series in which if n is odd, then an is 1/(n+[itex]\sqrt[]{n}[/itex]) while if n is even, then an is -1/n


    2. Relevant equations



    3. The attempt at a solution
    For 1., I tried integral test which seemed impossible to integrate, and then I tried comparison test but i can't find anything to compare to

    For 2. I thought that as n approaches infinity for 1/(n+[itex]\sqrt[]{n}[/itex]), which equals (1/[itex]\sqrt{n}[/itex])/([itex]\sqrt{n}[/itex]+1), then that expression pretty much looks like 1/n and thus the series is approx (-1)^n/n and is thus conditionally convergent like (-1)^n/n. But I then realized perhaps you can't apply the alternating series test since in the original series, the abs value of it is not decreasing for every term.E.g.1/(100+root(100))=1/110 < 1/101 Anyone have any idea on how to tackle this, or am I right from my original attempt?
     
    Last edited: Oct 4, 2013
  2. jcsd
  3. Oct 4, 2013 #2

    Zondrina

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    Let's start with #1.

    Your series is: ##\sum_{n=1}^{∞} \sqrt{n} - \sqrt{n}cos(\frac{1}{n})##

    I know this may seem misleading, but this series appears to converge.

    Hint: ##|cos(\frac{1}{n})| ≤ 1##
     
  4. Oct 4, 2013 #3

    haruspex

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    Yes, it converges, but you need to use a stronger fact about cos of small angles. Freshman2013, do you know the Taylor expansion of cos?
     
  5. Oct 4, 2013 #4

    jbunniii

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    For the first problem, this trig identity may be useful: ##2\sin^2(x) = 1 - \cos(2x)##.
     
  6. Oct 4, 2013 #5

    jbunniii

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    For problem 2, try adding ##a_n## and ##a_{n+1}##, where ##n## is odd.
     
  7. Oct 4, 2013 #6
    so for the second one, could I do this:
    the first term 1/(n+[itex]\sqrt{n}[/itex]) for all odd n, I rewrite as a separate series 1/(2n-1+[itex]\sqrt{2n-1}[/itex]) and the 2nd term -1/n for all even n I rewrite as -1/(2n)

    I combine those two series and I get (1-[itex]\sqrt{2n-1}[/itex])/(4n^2-2n+2n[itex]\sqrt{2n-1}[/itex]) and test for that(which is convergent). And I do the same except with 1/2n instead -1/2n to check for absolute convergence.
     
  8. Oct 4, 2013 #7

    jbunniii

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    For absolute convergence, you can do something simpler. I claim that if there is some subsequence ##(n_k)_{k=1}^{\infty}## such that ##\sum_{k=1}^{\infty} |a_{n_k}|## diverges, then ##\sum_{n=1}^{\infty} |a_n|## must also diverge. If you can prove this, then you can apply it immediately to your series.
     
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