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Finding the sum of following series

  1. Nov 4, 2009 #1
    1. The problem statement, all variables and given/known data

    n=3, where it goes to infinity
    [tex]\sum ln(n-2)[/tex]

    2. The attempt at a solution

    I'm not even sure how to start.. I just learned this today and I didn't understand this.. so I started solving questions from text then came across this question (didn't have an answer in the back)..

    am I suppose to use:

    n=0, where it goes to infinity
    [tex]\sum ar^n = a/(1-r)[/tex]
    but this equation starts from n=0..

    thank you for your help
     
  2. jcsd
  3. Nov 4, 2009 #2

    Mark44

    Staff: Mentor

    Your series can be expanded to
    ln(1) + ln(2) + ln(3) + ln(4) + ...
    Each term after the first is positive and getting larger. Is that enough of a start?
     
  4. Nov 4, 2009 #3
    i understand that part.. but I don't understand how I would use that to solve a question,
    how do I look for a final sum if the number keeps going on forever?
     
  5. Nov 4, 2009 #4
    I would say that you are summing up larger and larger numbers, not that the number keeps going on forever. Although, you are thinking of it correctly, but remember that all sums do not necessarily converge, that is they do not necessarily sum up to a finite number.
     
  6. Nov 4, 2009 #5

    Mark44

    Staff: Mentor

    An infinite series can either add up to a specific finite number (the series converges) or not (the series diverges). Divergent series can either add up to an infinitely large or infinitely negative value, or can oscillate between two or more values.
     
  7. Nov 4, 2009 #6
    oh i see..
    then how would I find if it converges or diverges?
    I think i remember prof cancelling a bunch of terms by dividing it by some number.. is this the way to do it?
     
  8. Nov 4, 2009 #7
    If the series
    [tex]\sum_{n=1}^\infty a_n[/tex]
    of real numbers converges, then [tex]\lim_{n\to\infty} a_n = 0[/tex].

    What is [tex]\lim_{n\to\infty} \ln(n-2)[/tex]?
     
  9. Nov 4, 2009 #8
    [tex]
    \lim_{n\to\infty} \ln(n-2)
    [/tex]

    wouldn't this be infinity?
     
  10. Nov 4, 2009 #9
    Exactly. So does this series converge? Look at the statement I posted.
     
  11. Nov 4, 2009 #10
    since it goes to infinity, it diverges.. I think..

    so for my question above, since it goes to invinity, it diverges also?
     
  12. Nov 4, 2009 #11
    Right. You are summing up larger and larger values that increase without bound. This means that the series will sum to infinity, and thus it is divergent. The contrapositive (i.e. an equivalent version) of my statement above is:

    If [tex]\lim_{n\to\infty} a_n \not=0[/tex], then the series
    [tex]\sum_{n=1}^\infty a_n[/tex]
    of real numbers diverges.
     
  13. Nov 4, 2009 #12
    ohh!!! wow thank you so much! I think I understand.. I thought all sums had a final number because I guess prof always used an example that gave number for result.. thank you again
     
  14. Nov 4, 2009 #13
    Be careful though. Just because the limit of the terms goes to zero, that doesn't necessarily mean that the series will converge. You should learn this soon. For example,
    [tex]\begin{alignat*}{2}
    \lim_{n\to\infty} \frac{1}{n} &= 0 &\quad\text{but}\quad \sum_{n=1}^\infty \frac{1}{n} &= \infty. \\
    \lim_{n\to\infty} \frac{1}{n^2} &= 0 &\quad\text{and}\quad \sum_{n=1}^\infty \frac{1}{n^2} &= \frac{\pi^2}{6}.
    \end{alignat*}[/tex]
     
  15. Nov 4, 2009 #14
    [tex]\quad \sum_{n=1}^\infty \frac{1}{n^2} &= \frac{\pi^2}{6}.
    \end{alignat*}[/tex]

    um how did you get pi^2/6..?
     
  16. Nov 4, 2009 #15
    I plugged it into Wolfram Alpha. See my http://www.wolframalpha.com/input/?i=sum+of+1/n^2+from+1+to+infinity". :)

    You won't have to worry about what these series converge to really, and there are ways to sum this. Although, I don't know any them off the top of my head. You can search this forum or Google to find ways, but I wouldn't worry about it right now.
     
    Last edited by a moderator: Apr 24, 2017
  17. Nov 4, 2009 #16

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    It's the Riemann zeta function of 2. You really figure it out by using contour integrals and residue theorem stuff, not Wolfram Alpha. n!kofeyn should really have given you an easy example of a convergent sequence like 1/2^n.
     
    Last edited by a moderator: Apr 24, 2017
  18. Nov 4, 2009 #17
    Well sorry. He asked how I knew it was pi2/6, so I told him. I mentioned there are ways to sum it up, but there is no point in mentioning those ways to a first or second semester calculus student, at least in the context of the original question. The point was to show how two very similar series, 1/n and 1/n2, differ in their convergence, and to illustrate the fact that I mentioned in my post. The point of that post and of the original question wasn't to show how you can sum up an easy series.

    You might as well mention you sum it up using rainbows and clouds if you're going to mention residues and contour integrals to a beginning calculus student.
     
  19. Nov 4, 2009 #18

    Mark44

    Staff: Mentor

    You should not use "it" here because you are talking about two different things. The first "it" refers to the general term in the series. The second "it" refers to the sum of all the terms.
     
  20. Nov 5, 2009 #19

    Gib Z

    User Avatar
    Homework Helper

    There's an elementary proof that most high school students can understand.

    http://en.wikipedia.org/wiki/Basel_problem

    Just for the heck of it, I'll give a nice proof [tex]\sum_{n=1}^{\infty} \frac{1}{n}[/tex] diverges.

    Start with the elementary inequality [tex]e^x > 1+x[/tex] for x>0. This can be reached in a few ways; truncating the expansion of the exponential functions taylor series, show the function f(x) = e^x - 1 -x is positive for x>0 and increasing, or perhaps most easily accessible to the poster, drawing a tangent to e^x at x=0.

    Consider, [tex]e^{\sum_{k=1}^n 1/k}= e^{1/1} e^{1/2} e^{1/3} e^{1/4} ... e^{1/n}[/tex]

    which by the inequality;

    [tex] > \left(1+ \frac{1}{1}\right)\left(1+ \frac{1}{2}\right)\left(1+ \frac{1}{3}\right)...\left( 1+ \frac{1}{n}\right)[/tex]

    which is equal to :

    [tex] \left( \frac{2}{1}\right)\left( \frac{3}{2}\right)\left( \frac{4}{3}\right)\left( \frac{5}{4}\right)...\left( \frac{n+1}{n}\right) = n+1[/tex]

    Since as n goes to infinity the lower bound diverges, [tex]e^{\sum 1/k}[/tex] also diverges and hence so does the sum.
     
  21. Nov 5, 2009 #20

    Dick

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    Science Advisor
    Homework Helper

    I gotcha. Sorry about the crabby tone of my note. It must have been late.
     
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