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Homework Help: Absolutely convergent or not?

  1. Dec 8, 2007 #1
    1. The problem statement, all variables and given/known data

    Decide if [tex]\sum_{n=1}^{\infty}(-1)^n\frac{\sqrt{n+1}-\sqrt{n}}{n} [/tex]

    is convergent and if it is, is it absolutely convergent or conditionally convergent?

    3. The attempt at a solution

    I'm pretty sure that the [tex] \lim_{n\rightarrow\infty} a_n = 0 [/tex]

    Am I supposed to use [tex] \frac{1}{n} [/tex] to compare with [tex] a_n [/tex] ?

    If I do that than no, it diverges, since 1/n does. But something tells me that is not correct.

    And the (-1)^n is confusing me a bit. I know that a_n is alternating because of it but is this telling me something? Can alternating series be absolutely convergent?

    Thank you

    Edit: I also tried to multyply denominator and numerator with sqrt(n+1)-sqrt(n)

    and I got [tex] \frac{1}{(n)(\sqrt{n+1}+\sqrt{n})} [/tex]

    If 1/n is smaller then the above. What does that tell me?
    Last edited: Dec 8, 2007
  2. jcsd
  3. Dec 8, 2007 #2


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    Alternating series are always convergent as long as the absolute value of the terms decreases monotonically and go to zero.

    To determine absolute convergence, you have to sum the absolute value of each term, essentially ignoring the (-1)n. This is why the last bit that you did is useful, because you should know something about whether
    [tex] \sum \frac {1}{n^s} [/tex] converges depending on s. Then note that what you got is smaller than [tex] \frac {1}{n^{3/2}} [/tex] for each term

    For a proof on alternating series: http://planetmath.org/encyclopedia/ProofOfAlternatingSeriesTest2.html [Broken]
    Last edited by a moderator: May 3, 2017
  4. Dec 8, 2007 #3
    Okey. The first line from your reply is great. It makes things more simple so I know that it is convergent.

    Now determing if it is absolute convergent. I have to see that [tex] \frac{1}{(n)(\sqrt{n+1}+\sqrt{n})} [/tex] is in fact smaller than [tex] \frac{1}{n^\frac{3}{2}} [/tex]

    Well there is a rule from my book that says that 1/n^p converges if p is bigger then 1.

    But this is where I get bit confused.

    Now I know that 1/n^(3/2) converges and is bigger than what I got. So is it absolutely convergent or what?

    What if 1/n^(3/2) would be smaller then what I got? Would that diverge to infinity then ?
    Thank you
  5. Dec 8, 2007 #4


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    If you see what you have is smaller than [tex] \frac{1}{n^\frac{3}{2}} [/tex], by the comparison test you're done. If you found that what you have is larger, then it still could be in between [tex] \frac{1}{n^\frac{3}{2}} [/tex] and [tex] \frac {1}{n} [/tex], so more analysis would be needed. But you should notice that [tex] \sqrt{n+1} + \sqrt{n} \geq \sqrt{n} [/tex] for all n
  6. Dec 8, 2007 #5
    Ok. So now I know that a_n is convergent and all I have to do is to do this again but with absolute values and see if I get the same result right? Because if [tex] a_n = /{a_n}/ [/tex] then it is absolut convergent.

    Sorry bout the "/" marks. I dont have the straight lines on my keyboard.

    Thank you Office Shredder, nice nick btw.
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