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Comparison test

  1. Feb 28, 2015 #1

    yuk

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    1. The problem statement, all variables and given/known data

    determine series convergence of divergence

    summation (n=1 to infinity) n/n^2 +1

    2. Relevant equations


    3. The attempt at a solution
    I take the limit comparison
    limit (1/n)/ (n/(n^2 +1) =1
    for 1/n if i use p series the series diverge
    if i use the method to take limit of sequence An then 1/n =0 so the series converge

    The answer is diverge
     
  2. jcsd
  3. Feb 28, 2015 #2

    LCKurtz

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    ##lim_{n\to\infty}a_n=0## does not imply convergence.
     
  4. Feb 28, 2015 #3

    yuk

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    but In the text book
    nTh term test
    if the sequence An converge to zero, then the series An converges

    I was so confusing
     
  5. Feb 28, 2015 #4

    SammyS

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    You should have the denominator in parentheses.

    n/(n^2 +1)

    You may be referring to the ratio test:

    Suppose ##\ \displaystyle \lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right| = r \, . ##
    Then:
    if r > 1, the series diverges
    if r < 1, the series converges
    if r = 1, the test is inconclusive. (appears to be the case here.)​

    Use the limit comparison test.
     
  6. Mar 1, 2015 #5

    Mark44

    Staff: Mentor

    That's not what it says.
    The nth term test for divergence says something along these lines.
    If a series ##\sum a_n## converges, then ##\lim_{n \to 0} = 0##

    The converse of this statement (i.e., if ##\lim_{n \to 0} = 0##, then the series ##\sum a_n## converges) IS NOT TRUE!! A classic example is the series ##\sum 1/n##. Even though ##\lim_{n \to \infty} 1/n = 0##, the series itself diverges.

    An equivalent way to state the nth term test for divergence is this:
    If ##\lim_{n \to 0} a_n \neq 0##, then the series ##\sum a_n## diverges.
     
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