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!

Finding Sum of the Series

  1. May 30, 2012 #1
    Ʃ (from n=1 to ınfınıty ) n /(n+1)!

    the questions asks sum of the this series.

    I try to write this series as Ʃ 1/(n-1)!.(n+1)but ı couldn't simulate any series.

    How can ı do this ?
     
  2. jcsd
  3. May 30, 2012 #2

    sharks

    User Avatar
    Gold Member

    Hi e179285

    [tex]\sum_{n=1}^{\infty} \frac{n}{(n+1)!}[/tex]
    Use the ratio test.
    [tex]u_n = \lim_{n\to\infty}\frac{n}{(n+1)!}=\lim_{n\to\infty}\frac{n}{(n+1)n!}
    \\u_{n+1} = \lim_{n\to\infty}\frac{(n+1)}{(n+2)!}=\lim_{n \to \infty} \frac{(n+1)}{(n+2)(n+1)n!}=\lim_{n \to \infty} \frac{1}{(n+2)n!}
    [/tex]
     
    Last edited: May 30, 2012
  4. May 30, 2012 #3
    Hi sharks,

    what ı can use ratio test for ?
     
  5. May 30, 2012 #4

    sharks

    User Avatar
    Gold Member

    Do you know the ratio test?

    Basically, it goes like this:
    [tex]\lim_{n \to \infty} \frac{u_{n+1}}{u_n}=L[/tex]If L < 1, the series converges. If L > 1, the series diverges, and if L = 1, then the result is inconclusive, meaning, the series may either converge or diverge, and you'll have to use another test to find out.
     
  6. May 30, 2012 #5
    ı know ratio test and ı found it is convergent but the question wants finding sum of the series I don't know how to do this
     
  7. May 30, 2012 #6

    sharks

    User Avatar
    Gold Member

    I misread your post. So, you want to find the limit of the series or the limit of the sequence of partial sums:
    [tex]\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+ \; ...
    \\\frac{1}{2}+\frac{2}{6}+\frac{3}{24}+ \; ...
    \\\frac{1}{2}+\frac{1}{3}+\frac{1}{8}+ \; ...[/tex]
     
  8. May 30, 2012 #7
    Try applying the Taylor series for derivative of [tex]f(x) = \frac{e^x}{x}[/tex] at x=1.
     
  9. May 30, 2012 #8

    ehild

    User Avatar
    Homework Helper
    Gold Member

    Is not it rather the derivative of [tex]f(x) = \frac{e^x-1}{x}[/tex] at x=1?

    ehild
     
  10. May 30, 2012 #9
    thanks for answers :)
     
  11. May 30, 2012 #10
    I believe you can do it both ways. If you use [tex]f(x) = \frac{e^x}{x}[/tex] then you get a constant(not giving the value here, because it's the answer!) plus the series of which the sum needs to be found. Subtracting it both the sides would give the sum.

    If you use [tex]f(x) = \frac{e^x-1}{x}[/tex] then that extra term gets subtracted in the expansion, and is compensated by the differentiation on LHS, because the function is framed that way.
     
  12. May 30, 2012 #11

    sharks

    User Avatar
    Gold Member

    I'm curious to know how that function for the Taylor series at x=1 was obtained in the first place? Is there a theorem relating those two specific results?
     
  13. May 30, 2012 #12
    The Taylor series was just an idea to relate the given series, because you see there are factorial terms in the denominator pretty much laid out the way you need them. There was just a need for manipulation to give you the exact series, according to which the function was decided. A different series sum could be found by possibly using [itex]f(x)=sin(x)/x^2[/itex], it all depends on the given problem. As far as I know, there is no 'theorem' relating the two.

    I think the above sum can be evaluated by yet another method(not involving Taylor series), but I haven't been able to complete the solution yet.
     
  14. May 30, 2012 #13

    ehild

    User Avatar
    Homework Helper
    Gold Member

    You know the Taylor series of ex: The terms have n! in the denominator. [tex]e^x=\sum_0 ^{\infty} \frac{x^n}{n!}[/tex]Integrating it, you get n+1! in the denominators:

    [tex]\int_0^x{e^udu}=e^x-1=\sum_0 ^{\infty} \frac{x^{n+1}}{n+1!}=x+\sum_1 ^{\infty} \frac{x^{n+1}}{n+1!}[/tex]
    Divide both sides by x:

    [tex]f(x)=\frac{e^x-1}{x}=1+\sum_1 ^{\infty} \frac{x^{n}}{n+1!}[/tex]

    Take the derivative of both sides:

    [tex]f'(x)=\sum_1 ^{\infty} \frac{n x^{n-1}}{n+1!}[/tex]

    Substituting x=1, you get the series.
     
  15. May 30, 2012 #14

    ehild

    User Avatar
    Homework Helper
    Gold Member

    The derivative of ex/x is zero at x=1.

    ehild
     
  16. May 30, 2012 #15
    Yes it is, but,

    [tex]f'(x)= \frac{-1}{x^2} + 0 + \frac{1}{2!} + \frac{2x}{3!} + \frac{3x^2}{4!}+.....[/tex]

    Putting x=1, gives you the answer.

    Actually, after your idea I realized you can use Taylor series for any

    [tex]f(x) = \frac{e^x - a}{x}[/tex]

    at x=1, where a is constant.
     
  17. May 30, 2012 #16

    ehild

    User Avatar
    Homework Helper
    Gold Member

    Well, but it is not a Taylor series and not the derivative of a Taylor series.

    ehild
     
  18. May 30, 2012 #17
    A quick way to evaluate this without using Taylor polynomials, e^x, etc is to make it into a telescoping series. n/(n+1)!= n/(n+1)! -1/(n+1)! + 1/(n+1)! = 1/n! -1/(n+1)! ... Plug in the first few terms to see only the first term and the last term remains... Since your last term goes to 0, your first term in your answer
     
  19. May 30, 2012 #18
    Here is the elegant method! :approve:


    As I know it, Taylor series expansion about a point x=a is given by

    [tex]f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)(x-a)^2}{2!} + .....[/tex]

    Using that for ex at a=0, we get the expansion,

    [tex]e^x = e^0(1+ (x-0) + \frac{(x-0)^2}{2!} + .....)[/tex]

    Which I used in the above post.
     
  20. May 30, 2012 #19

    ehild

    User Avatar
    Homework Helper
    Gold Member

    Excellent! :smile:

    ehild
     
  21. May 30, 2012 #20

    ehild

    User Avatar
    Homework Helper
    Gold Member

    Yes, that is why a series with negative power of x (or x-a) is not a Taylor series .

    But your Taylor expansion method was a splendid idea!


    ehild
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Finding Sum of the Series
Loading...