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Posers and Puzzles

Posers and Puzzles

  1. 09 Apr '08 20:49
    How do I figure out what this series converges to?:

    Sum: 1/[n! (n+2)], n goes from 0 to infinity.

    In case there's any confusion, this is (n+1)/[(n+2)!]
  2. 09 Apr '08 21:56
    Originally posted by Jirakon
    How do I figure out what this series converges to?:

    Sum: 1/[n! (n+2)], n goes from 0 to infinity.

    In case there's any confusion, this is (n+1)/[(n+2)!]
    I think it's 1.

    Too late to type up the answer now, so here's a clue. Consider:

    f(x) = SUM{n = 0, inf} x^(n+2)/[n!(n+2)]
  3. 09 Apr '08 22:16
    Originally posted by mtthw
    I think it's 1.

    Too late to type up the answer now, so here's a clue. Consider:

    f(x) = SUM{n = 0, inf} x^(n+2)/[n!(n+2)]
    I thought it would be converging on zero... just due to the fact that as n gets infinitely large the bottom of the fraction is going to be exponentially larger than the top thus bringing you closer and closer to zero...
  4. Standard member adam warlock
    Baby Gauss
    09 Apr '08 23:07
    Originally posted by strokem1
    I thought it would be converging on zero... just due to the fact that as n gets infinitely large the bottom of the fraction is going to be exponentially larger than the top thus bringing you closer and closer to zero...
    What converges to 0 is the expression of the general term of the series while n goes to infinity. But as you take more and more terms the series approaches a fixed value since it is obvious that that series converges.

    For instance 1+1/2+1/4+1/8+1/16+1/32+...+1/2^n+...=2. What goes to 0 is 1/2^n that is the general term.

    Actually the fact that the general term goes to 0 as n goes to infinity is a necessary but not sufficient condition for a series to converge.
  5. 10 Apr '08 00:45
    I think it's 1.

    I know it's 1, but I don't know how.
  6. Standard member TheMaster37
    Kupikupopo!
    10 Apr '08 06:21
    Originally posted by adam warlock
    What converges to 0 is the expression of the general term of the series while n goes to infinity. But as you take more and more terms the series approaches a fixed value since it is obvious that that series converges.

    For instance 1+1/2+1/4+1/8+1/16+1/32+...+1/2^n+...=2. What goes to 0 is 1/2^n that is the general term.

    Actually the fact that the ...[text shortened]... to 0 as n goes to infinity is a necessary but not sufficient condition for a series to converge.
    Over here we have two seperate terms for convergence of series.

    In this case the series itself converges to 0, this is obvious.

    The SUM-series however converges to something something else.
  7. 10 Apr '08 09:14
    Originally posted by TheMaster37
    Over here we have two seperate terms for convergence of series.

    In this case the series itself converges to 0, this is obvious.

    The SUM-series however converges to something something else.
    With the conventions I was taught, a series always refers to a sum of a sequence. The sequence converges to zero.

    Wikipedia agrees - it must be true!
  8. 10 Apr '08 09:18 / 1 edit
    Originally posted by Jirakon
    I think it's 1.

    I know it's 1, but I don't know how.
    The clue didn't help then?

    If we write f(x) = SUM x^(n+2)/[n!(n+2)], note that f(1) gives the series you want.

    df/dx = SUM x^(n+1)/n! = x(SUM x^n/n!)

    You might recognise last sum: SUM x^n/n! = e^x

    so df/dx = xe^x
    => f = (x - 1)e^x + c

    To calculate c, set x = 0. We know f(0) = 0, so c = 1

    => f(x) = (x - 1)e^x + 1
    => f(1) = 1

    One of those cases where the easiest way to solve a problem is actually to solve a more general problem, which has a special case that matches your original problem.
  9. Standard member adam warlock
    Baby Gauss
    10 Apr '08 10:18
    Originally posted by TheMaster37
    Over here we have two seperate terms for convergence of series.

    In this case the series itself converges to 0, this is obvious.

    The SUM-series however converges to something something else.
    I was thought that the word series means to the actual value of the limiting sum. And that what is being summed is called "termo geral da série". I translated that to english as general term but maybe that isn't the correct technical term in english. Anyway this general term must tend to 0 for the series to converge. Now if the series converge to 0 either all terms are 0 or some kind of cancelling must occur between succesive terms.
  10. Subscriber deriver69
    Keeps
    10 Apr '08 10:18
    I think I would use
    (n+1)/(n+2)! =(n+2-1)/(n+2)!
    =(n+2)/(n+2)!-1(n+2)!
    =1(n+1)!-1/(n+2)!


    Use the method of differences and you are left with 1/1!-a successively small term which tends to 0 as the series has an infinite number of terms
  11. 10 Apr '08 10:52
    Originally posted by deriver69
    I think I would use
    (n+1)/(n+2)! =(n+2-1)/(n+2)!
    =(n+2)/(n+2)!-1(n+2)!
    =1(n+1)!-1/(n+2)!


    Use the method of differences and you are left with 1/1!-a successively small term which tends to 0 as the series has an infinite number of terms
    Ah yes, that's simpler than my method.

    Having another look at the question, I think the way it's stated makes a difference. With the first formula, the idea of differentiating to get rid of the (n + 2) term leapt out at me. If I'd used the second (equivalent) formula I think I'd have been more likely to take your approach.