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Help finding Whether Series Converges or Diverges

  1. Oct 14, 2013 #1
    1. The problem statement, all variables and given/known data
    Find whether the following converge or diverge

    A:[itex]\sum_{n=1}^{\infty} \frac{n-1}{n}[/itex]

    B:[itex]\sum_{n=1}^{\infty} \frac{1}{n^{\frac{3}{2}}}[/itex]



    2. Relevant equations



    3. The attempt at a solution
    Again really not sure how to show this mathematically, not even sure if what I have done is correct so any help or advice is much appreciated.

    A:
    [tex]
    \sum_{n=1}^{\infty} \frac{n-1}{n}=\frac{\frac{n}{n}-\frac{1}{n}}{\frac{n}{n}}=1-\frac{1}{n} \\
    =1-1+1-\frac{1}{2}+1-\frac{1}{3}+1-\frac{1}{4}...[/tex]
    here it seems it diverges to infinity as n tends to infinity (due to the +1's)

    B:
    [tex]
    \sum_{n=1}^{\infty} \frac{1}{n^{\frac{3}{2}}}=1+0.353+0.192+.....[/tex]
    Here is looks like it converges to the limit of 2 as n tends to infinity.
     
  2. jcsd
  3. Oct 14, 2013 #2

    Ray Vickson

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    What are some of the necessary conditions that the terms of a convergent series must satisfy? Do the terms in A satisfy those conditions? As for B: what tests have you learned for determining convergence? What happens if you apply these tests?
     
  4. Oct 14, 2013 #3
    He showed us very quicly something called the ratio test on Friday. But there is a seperate problem ont he worksheet which specifically asks to do the ratio test. Sorry if it looks as though I have not put my best effort into this but I have, I spent two to three hours on all of it last night and this is the ebst I can come up with.
     
  5. Oct 14, 2013 #4
    I think I may have made some progress on my understanding what is needed by the question.

    A:
    [tex]\sum_{n=1}^{\infty} \frac{n-1}{n}=\frac{\frac{n}{n}-\frac{1}{n}}{\frac{n}{n}}=1-\frac{1}{n} [/tex]
    So as it equals 1-1/n as n gets bigger that fraction will get closer to 0 so this must converge to the limit of 1?

    B:
    [tex]\sum_{n=1}^{\infty} \frac{1}{n^{\frac{3}{2}}}[/tex]
    With this one, as n gets bigger the denominator gets bigger and therefore the fraction gets smaller so this one must converge to the limit of 1?
     
  6. Oct 14, 2013 #5

    Office_Shredder

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    For A, there is a significant difference between
    [tex] \lim_{n\to \infty} 1 - \frac{1}{n} [/tex]
    and
    [tex] \sum_{n=1}^{\infty} 1-\frac{1}{n} [/tex]

    You are correct that the terms of your sum are converging to 1, but this does not mean that the sum itself is converging to 1. One of the first properties of infinite series you learned should have been a statement around the lines of "If [itex] \sum a_n [/itex] is convergent, then [itex] \lim_{n\to \infty} a_n [/itex] is something specific" (where 'something specific' is omitted here to avoid stating the whole answer to the problem).
     
  7. Oct 14, 2013 #6
    Honestly I cant find anything similar to that text you have quoted in the lecture notes, this was a very quick thing we had 3 50 minute lecture on the topic and then were given this problem sheet. We started something new today.

    But using rationale, if the terms are converging to 1, and the sum is all the terms added up, then surely it diverges to infinity? As 1+1+1.... is infinity?
     
  8. Oct 14, 2013 #7

    Office_Shredder

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    Yes, this is the main point. OK the formal statement is that if the series converges then
    [tex] \lim_{n\to \infty} a_n = 0[/tex]
    necessarily (so if the limit is not equal to that, then the series diverges)

    Note that this doesn't work both ways - if the terms go to zero, the sum itself may still diverge so you can't use this on part B.
     
  9. Oct 14, 2013 #8
    OK thanks.

    Does part B also diverge though, even though the terms themselves get closer to 0?
     
  10. Oct 14, 2013 #9

    jbunniii

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    The terms converging to zero doesn't guarantee convergence or divergence. For example,
    $$\sum_{n=1}^{\infty} \frac{1}{n}$$
    diverges, but
    $$\sum_{n=1}^{\infty} \frac{1}{n^2}$$
    converges, even though in both cases the terms converge to zero. You will need to apply a convergence test to determine the fate of ##\sum_{n=1}^{\infty} \frac{1}{n^{3/2}}##. Which tests do you know?
     
  11. Oct 14, 2013 #10
    I only know of the ratio test and the comparison test, but dont fully understand either of them.

    This is my go of using the ratio test on it.

    [tex]
    \frac{a_{n+1}}{a_n}
    [/tex]
    So it will become
    [tex]
    \frac{\frac{1}{(n+1)^{1.5}}{\frac{1}{n^{1.5}}} = \frac{1}{(n+1)^{1.5}} * \frac{1}{n^{1.5}} \\
    = \frac{n^{1.5}}{(n+1)^{1.5}}
    [/tex]
    But thats as far as I got and am unsure as to what to do with it now.

    With the comparison test I dont understand where the b_n comes from:
    [itex]0 <= a_n <= b_n[/itex] for all n and then the if the sum to infinity of a_n diverges then b_n diverges, and if b_n converges, a_n converges. I just dont get where the b_n comes from.

    EDIT: I have gone through that latex code several times and I cant seem to find the error.
     
    Last edited: Oct 14, 2013
  12. Oct 14, 2013 #11

    Office_Shredder

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    The comparison test is used when you a complicated expression in your sum, but you know how to deal with a simpler one that looks similar... it's not really appropriate here.

    The ratio test will not work either (try re-writing that fraction as [itex] \left( \frac{n}{n+1} \right)^{3/2} [/itex] ).

    If you are really supposed to be able to answer part (b) at this point in your class you must have learned something called the integral test, you should take a look back in your notes to see if it's there.
     
  13. Oct 14, 2013 #12
    We definitely have not done the integral test, I would have remembered plus my lecturer does a summary of each week and last week was this topic (this weeks it is functions) and it would be on there but there is only the ratio and comparison test.

    I can look it up and give it a go though.
     
  14. Oct 14, 2013 #13

    Office_Shredder

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    Are you following a textbook which contains extra material not covered in the lectures perhaps? Because the questions you are asking do not correlate with the limited amount of information you say you've gotten in your lectures - basically I don't understand how you are supposed to answer them. Maybe you just were told something like a "p-test" which says when
    [tex] \sum_{n=1}^{\infty} \frac{1}{n^p} [/tex]
    converges or diverges
     
  15. Oct 14, 2013 #14
    Ah yes, we have been shown that. However looking at it, as the starting index is 1, that is inconclusive, but for the rest of the term p > 1, so does that mean it diverges to infinity? This is a worksheet given out by our lecturer which is assessed and will go towards our final grade for the unit.

    I have just briefly looked at the integral test as well, and came up with the same answer (that it diverges to infinity. Does the following make sense?

    Its confirmed that it is always positive, and is obvious that it is decreasing .
    [tex]
    \int_1^{\infty} \frac{1}{x^{1.5}} = \lim_{t \to \infty} \int_1^{t} \frac{1}{x^{1.5}} = \lim_{t \to \infty} \ln x^{1.5} \\
    \lim_{t \to \infty} \ln t^{1.5}
    [/tex]
    (I do not know the latex commands for that tall bar with the limits on is)

    And as ln x is infinity when x tends to infinity, then limit of ln t^1.5 will also be infinity as t tends to infinity.

    And as the integral has no limit, then the series diverges.
     
  16. Oct 14, 2013 #15

    Mark44

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    Your integration is incorrect.
    ##\int \frac{dx}{x^{1.5}} \neq ln(x^{1.5})##
     
  17. Oct 14, 2013 #16

    Office_Shredder

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    Can you say precisely what your test says? I don't understand your description of why you think it diverges
     
  18. Oct 14, 2013 #17
    Sorry. I meant converges. I was reading the lecture notes in a rush and misread it. If p > 1 is converges and if p <= 1 then is diverges.
     
  19. Oct 14, 2013 #18

    Office_Shredder

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    Yes, that's correct.
     
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