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Homework Help: Which one of the following sequences converge

  1. Feb 5, 2005 #1
    Hello, could someone please help me with the following questions?

    Q. Determine which one of the following sequences converge and which do not converge. Explain your answers. For any sequence that converges, find the limit.

    (i) [tex]\frac{{n + \left( { - 1} \right)^n }}{{2 + \left( { - 1} \right)^n }}[/tex]

    (ii) [tex]e^{ - n} n^5 \log _e \left( n \right)[/tex]

    I am not sure about part (i). As far as I know, (-1)^n diverges, so doesn't that mean that [tex]\frac{{n + \left( { - 1} \right)^n }}{{2 + \left( { - 1} \right)^n }}[/tex] also diverges? If not can someone please explain why to me?

    For part (ii) I think it does coverge and the limit can be found by repeatedly using L'Hospital's rule. Am I on the right track?

    Any help is appreciated.
  2. jcsd
  3. Feb 5, 2005 #2
    For i), find a divergent subsequence.
  4. Feb 5, 2005 #3
    Could you please tell me what a sub sequence is? I want to make sure so that my limited understanding of sequences does not lead me to misinterpret your advice.
  5. Feb 5, 2005 #4


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    Suppose you've got a sequence [tex]x_{n}[/tex].
    Let f=n(k) be an integer function of k, and that f is a strictly increasing function of k.

    Then, you may define a new sequence:
    We say that [tex]a_{k}[/tex] is a subsequence of the x-sequence, for natural reasons.
    As an example, [tex]a_{k}=x_{2k}[/tex] is a subsequence of x consisting of the even members of x.
    Last edited: Feb 5, 2005
  6. Feb 5, 2005 #5


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    Or use:

    [tex]\frac{{n + \left( { - 1} \right)^n }}{{2 + \left( { - 1} \right)^n }} \geq \frac{n-1}{3}[/tex]
    for all natural n.
  7. Feb 5, 2005 #6


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    This time it has nothing to do with your exercise.Remember that a sequence diverges in 2 cases:
    1.If it doesn't have a limit ((-1)^{n} is a good example).
    2.It has a limit which is either + or - infty.

    In this case,(-1)^{n} and its divergence do not play any role...

  8. Feb 5, 2005 #7
    Thank you for your valuable time and help everyone. I should be able to finish this question now.
  9. Feb 6, 2005 #8
    Sorry I have some more questions :biggrin: . The alternating series test as stated in my book is as follows.

    If the alternating series [tex]\sum\limits_{n = 1}^\infty {\left( { - 1} \right)^{n - 1} b_n = } b_1 - b_2 + b_3 - b_4 + b_5 - b_6 + ...,b_n > 0[/tex] satisfies:

    (a) [tex]b_{n + 1} \le b_n [/tex]

    (b) [tex]\mathop {\lim }\limits_{n \to \infty } b_n = 0[/tex]

    then the series is convergent.

    One of the examples in my book is: [tex]\sum\limits_{n = 1}^\infty {\frac{{\left( { - 1} \right)^n 3n}}{{4n - 1}}}[/tex]

    They take the limit of b_n which turns out to be (3/4). Since the limit is not equal to zero then isn't the sequence is not convergent and therefore divergent? I ask this because the book then goes on to show that the limit [tex]\mathop {\lim }\limits_{n \to \infty } \frac{{\left( { - 1} \right)^n 3n}}{{4n - 1}}[/tex] does not exist and concludes that the series is divergent as a consequence of the 'Test for Divergence.'

    My question is, if the conditions( (a) and (b) ) for an alternating series to be convergent are not met, can it be deduced that the series is divergent. Or does further work need to be done(like the book does) to do so?

    Also how come [tex]\mathop {\lim }\limits_{n \to \infty } \frac{{\left( { - 1} \right)^n 3n}}{{4n - 1}}[/tex] does not exist where as [tex]\mathop {\lim }\limits_{n \to \infty } \frac{{\left( { - 1} \right)^n }}{n}[/tex] does? Does it have anything to do with the fact that the second limit is equal to zero whereas the first one is not? I've seen many variations of limits involving (-1)^n in my book, some eixst, others do not. So I am really confused about it. Could someone please help me out by explaining or alluding to why the first limit I referred to in this paragraph does not exist whereas the second one does?
    Last edited: Feb 6, 2005
  10. Feb 6, 2005 #9


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    That's correct. If the conditions for the alternating series test is not met, then nothing can be said about the convergence or divergence of the series.
    Therefore some other method must be found. In this case, since the terms do not approach zero, the series is divergent.

    In general, you need more work.
    But in case condition (b) is not met, the sequence of terms does not approach zero and thus the series is divergent by the 'Test for Divergence'.

    Try to answer this one yourself. Consider the subsequence for odd n, and the one for even n and see where they converge to. What can you conclude about the original sequence.
  11. Feb 6, 2005 #10


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    I have strong reasons to believe that:

    [tex] \sum_{n=1}^{+\infty} \frac{(-1)^{n+1}}{n} =\ln 2 \approx 0.693 [/tex]

  12. Feb 6, 2005 #11
    dextercioby - My book seems to say the same thing, apart from how they have (n-1) where you have (n+1). I'm not sure if it makes a difference though as I am quite new to this topic.

    Galileo - The second sequence seems to converge to zero for both even and odd values of n whereas the first sequence does not appear to have that same property.

    Thanks again for the help.
  13. Feb 6, 2005 #12


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    Expand both formulas & compare the first 4-5 terms...Are they the same...??

  14. Feb 7, 2005 #13


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    If the sequence for even n converges to 3/4 and the one for odd n converges to -3/4, then the original sequence cannot converge. It will alternate between values close to 3/4 and -3/4.

    [itex]\frac{(-1)^n}{n}[/itex] does converge, because the sequence of absolute values of the terms
    goes to zero as n goes to to infinity.

    Also, if you know that the series [itex]\sum a_n[/itex] converges, you also know that [itex]a_n \to 0[/itex] as [itex]n\to \infty[/itex]. In your case, since the sequence tends to [itex]\ln 2[/itex] it converges, so the sequence of terms go to zero.
    Last edited: Feb 7, 2005
  15. Feb 7, 2005 #14
    dextercioby - Yeah it turns out that there is no difference. :biggrin:

    Galileo - Thanks for your post, it verified what I had in mind after I did what you suggested to in your previous post.
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