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Homework Help: Convengence of gemoetric squence

  1. Jul 16, 2010 #1
    1. The problem statement, all variables and given/known data

    Determine whether the sequence converges or diverges.


    2. Relevant equations

    3. The attempt at a solution

    In my note it appeared that "the sequence goes to 0, now we investigate the convergence of geometric series..." which I found it questionable today.

    I thought that if r is between -1 and 1, the limit ar^(n-1) = 0, and this says convergence
    in this case, our r is still (5/3) isn't it? So 5/3 is larger than 1, this means lim ar^(n-1) does not exist, which says divergence.

    Please verify that for me.

    Convergence of a geometric sequence
    Last edited: Jul 16, 2010
  2. jcsd
  3. Jul 16, 2010 #2


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    Yes. 5/3>1 so the limit of the terms does not exist and the sum of the series does not exist.
  4. Jul 16, 2010 #3
    Hi Dick. Thank you.
    Here is another question.

    For an infinite series, the series is said to converge to L if the sequence of its nth partial sum converges to L. Otherwise, the series diverges.

    In the case of a telescoping series, if I find that the nth partial sum goes to 5/12 after computing the limit, according to the definition of infinite series, the series will also converges to 5/12.

    But the Kth test for divergence states that: if limit ak =/= 0, then the series must not converges, which does not agree with the definition of an infinite series.

    I see that for a geometric series, the r has to be between -1 and 1 in order to be converges (which limit ak = 0).

    Moreover, 1/k is divergent due to it is a harmonic series (and also a typical p-series with p <= 1), whereas the Kth test suggest that it does converge because limit 1/k = 0)

    Indeed, that kth test suggest that if it does = 0, it may or may not converge, so I see that we need another approach for cases like 1/k which i know already. but whatever the case with =/= 0 and still converge?

    So when do I use the k-th test theorem?
  5. Jul 16, 2010 #4


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    The Kth test is talking about the limit of the individual terms in the series, not the partial sums. If you look at your telescoping series you have better have limit ak=0.
  6. Jul 16, 2010 #5
    Thank you for your reply.
    But the definition of infinite series states if the partial sum does converge, the series itself will also converges.

    I know that we take the limit of Sn (partial sum) of the telescoping series, and find the L. If L exists, that L is the what the series itself will converge too.

    Similarity,the definition of Kth test says if the limit of ak =/= 0, the series must be divergent.

    What I don't get is that since series is sum of a sequence. Ak is a sequence, which we denote series as sum of ak from n =k, to infinity. So how come the kth test fails in the case with special case like telescoping L = 5/12?

    The partial sum indeed goes to 5/12, and each of the term, gets smaller and smaller.


    But L is not zero!

    If we argue that 1/n, n >= 1, then it's sequence is 1, 1/2, 1/3, 1/4, and so on.
    I know that kth suggests that since 1/n goes to zero, it MAY, OR MAY NOT converges. That's another story. But each individual term does get smaller and smaller. If we compute the sum it wouldn't go to zero, obviously. There is no alternation. The nth partial sum of 1/k probably goes to a particular number that is between 1.5 <= w <= 2
    Last edited: Jul 16, 2010
  7. Jul 16, 2010 #6


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    The sum of the series is 5/12. The limit of the partial sums is the same thing. The series converges. ak=1/(k^2+4*k+3) and lim ak as k->infinity=0. I don't see any conflict. Are you sure you are reading the statement of this 'kth test' correctly? What does it say exactly?
  8. Jul 16, 2010 #7
    Hi, Thank you for ur reply.

    You pointed out the confusion that I had.

    ak=1/(k^2+4*k+3) goes to zero. Right. It suggests that it may or may not converge. This does not says the series will go converge to zero.

    Am I correct? If so I must have confused myself.

  9. Jul 16, 2010 #8


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    Right. ak->0 says the series MAY converge (but not necessarily to zero). If ak does NOT go to zero then the series definitely does NOT converge.
  10. Jul 16, 2010 #9
    Thank you Dick.
  11. Jul 16, 2010 #10


    Staff: Mentor

    The k-th term test (AKA the k-th term test for divergence) doesn't say anything about [itex]\sum a_k[/itex] if lim ak = 0. What it says is that if lim ak [itex]\neq[/itex] 0, then [itex]\sum a_k[/itex] diverges.

    Example 1: [itex]\sum 1/k[/itex] diverges
    Example 2: [itex]\sum 1/k^2[/itex] converges

    For both series, lim ak = 0.
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