1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Series equation

  1. Jan 16, 2009 #1
    1. The problem statement, all variables and given/known data
    I have to find the number of (real) solutions to the following equation and place them in intervals of size 1.
    [tex]\sum_{n=1}^{\infty}\frac{x^n+n^2+n}{x^{n+1}\cdot n\cdot(n+1)}=1[/tex]


    2. Relevant equations



    3. The attempt at a solution
    I have been able to prove that x > 1, because the series does not converge if x < 1. I was thinking of finding how much the series is, but I don't think I can do that (or at least I don't know how to).
     
    Last edited: Jan 17, 2009
  2. jcsd
  3. Jan 16, 2009 #2

    Mark44

    Staff: Mentor

    This isn't an equation, so it doesn't make sense to look for solutions.

    Are you asking for what values of x (as intervals) this series converges?
     
  4. Jan 16, 2009 #3
    Yes, that's what I'm asking. Thanks for any help.
     
    Last edited: Jan 16, 2009
  5. Jan 16, 2009 #4

    Mark44

    Staff: Mentor

    Can you show what you did to establish that the series diverges for x < 1 and that it converges for x > 1?
     
  6. Jan 17, 2009 #5
    If x < 1:
    [tex]\lim_{n\rightarrow\infty}\frac{x^n+n^2+n}{x^{n+1}\cdot n\cdot(n+1)}=\lim_{n\rightarrow\infty}\frac{n\cdot(n+1)}{x^{n+1}\cdot n\cdot(n+1)}=\lim_{n\rightarrow\infty}\frac{1}{x^{n+1}}=\infty[/tex]
    So the series does not converge.

    However if x > 1:
    [tex]\lim_{n\rightarrow\infty}\frac{x^n+n^2+n}{x^{n+1}\cdot n\cdot(n+1)}=\lim_{n\rightarrow\infty}\frac{x^n}{x^{n+1}\cdot n\cdot(n+1)}=\lim_{n\rightarrow\infty}\frac{1}{x\cdot n\cdot(n+1)}=0[/tex]
    So the series might converge if x > 1.

    [tex]\lim_{n\rightarrow\infty}n\cdot\left(1-\frac{a_{n+1}}{a_{n}}\right)=\lim_{n\rightarrow\infty}n\cdot\left(1-\frac{\frac{1}{x\cdot (n+1)\cdot(n+2)}}{\frac{1}{x\cdot n\cdot(n+1)}}\right)=\lim_{n\rightarrow\infty}n\cdot\left(1-\frac{n}{n+2}\right)=\lim_{n\rightarrow\infty}\frac{2n}{n+2}=2[/tex]
    Therefore the series converges if x > 1.
     
    Last edited: Jan 17, 2009
  7. Jan 17, 2009 #6
    Oh I just noticed for some reason the tex didn't display the equation properly...
    I want to find the solutions so that the series equals to 1, but it didn't display it.
    [​IMG]
    Sorry about that.
     
  8. Jan 17, 2009 #7

    Mark44

    Staff: Mentor

    What's your justification for the expression after the first = sign? Why does x disappear from the numerator, but not the denominator? If you're cancelling, you can't do that, since the x term isn't a factor in the numerator.
    What's your justification for the expression after the first = sign? Why did the terms in n disappear from the numerator?
     
  9. Jan 17, 2009 #8

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Sure the series converges for x>1. Break it up into 1/(x*n*(n+1))+1/x^(n+1). Compute the sum of each series. One is geometric and one telescopes.
     
  10. Jan 17, 2009 #9
    If x < 1, xn is 0 as n approaches infinity whereas n(n+1) approaches infinity, so we can take xn out.
    If x > 1, xn approaches infinity much faster than n(n+1), so the following is true: xn + n(n+1) ~ xn.

    OK, so I did split into geometric and telescope, added them back and found 1/(x-1) = 1. So x = 2, is that correct?
     
  11. Jan 17, 2009 #10

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    That's what I got.
     
  12. Jan 18, 2009 #11
    Thanks a lot for your help!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?