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Proving absolute converge with minorant criterion

  1. Apr 8, 2016 #1
    1. The problem statement, all variables and given/known data

    Hi everybody! I'm stuck on a problem in which I don't manage to prove absolute convergence (or not) of the series ∑ (-1)j/(√j + √(j+1)).

    2. Relevant equations

    Leibniz criterion, minorant criterion, limit comparison test, maybe others

    3. The attempt at a solution

    So first I noticed that the series alternates, so I ran the Leibniz test in order to find out if the series converges/diverges:

    |an+1| - |an| = 1/(√(j+1) + √(j+2)) - 1/(√j + √(j+1))
    = (√j - √(j+2))/[(√(j+1) + √(j+2))⋅(√j + √(j+1))] < 0
    ⇒ the sequence is monotone decreasing.

    lim j→∞ (1/(√j + √(j+1))) = 0
    ⇒ the series converges.

    Now that was rather easy, but I know that in order to prove that the series absolutely converges or not I must find out if the sequence |an| converges or diverges.

    And that's my problem. I imagine it being divergent, but I cannot seem to find a known divergent series smaller than 1/(√j + √(j+1)) (which btw can be rewritten as (√(j+1) - √j)). Any hint?

    Maybe I'm just using the wrong criterion. I just tried the limit comparison test, which didn't work, but that might be because I never used it (somehow we don't have this criterion in our class though it seems very important!!). Any advices about that test? How do you choose your "other series" to run the test with?


    Thank you very much in advance for your answers.


    Julien.
     
  2. jcsd
  3. Apr 8, 2016 #2
    Actually I just got an idea: if I factor the √j out, I get:

    √(j+1) - √j = √j (√(1+(1/j)) - 1)

    which is bigger than √j for all j ≥ 1. √j diverges, therefore √(j+1) - √j diverges too so the original series converges but not absolutely.

    Does that make sense? Any advices about the strategies to use for majorant/minorant criterions?

    EDIT: Oh no that was wrong... It's actually smaller than √j! :(
     
  4. Apr 8, 2016 #3

    PeroK

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    After your first step, what about looking at partial sums?
     
  5. Apr 8, 2016 #4

    Ray Vickson

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    1. You can do the sum ##S_n = \sum_{i=1}^n [\sqrt{i+1} - \sqrt{i}\:]## explicitly, to obtain a very simple formula for ##S_n## in terms of ##n##.
    2. You can, instead, use the fact that ##1/[\sqrt{i+1}+\sqrt{i}] > 2/\sqrt{i+1}##, then use the fact that the "p-series" ##\sum 1/n^p## converges if and only if ##p > 1##. This last fact has been discussed many times in this Forum. Alternatively, you can use the fact that ##1/\sqrt{i+1} > 1/(i+1)## and then use the fact that the harmonic series ##1/n## is divergent.
     
    Last edited: Apr 8, 2016
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