1. Limited time only! Sign up for a free 30min personal 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!

Proving a limit (x^n/(n!))

  1. May 29, 2016 #1
    1. The problem statement, all variables and given/known data

    Section is on using power series to calculate functions, the problem is on proving the limit, solution is also attached but I do not see how the solution proves the limit.

    2. Relevant equations

    Convergent power series form

    3. The attempt at a solution

    I attempted to represent the series with a Maclaurin series, but that leads nowhere. Could someone help explain why the solution given works?
     

    Attached Files:

  2. jcsd
  3. May 29, 2016 #2

    Mark44

    Staff: Mentor

    The problem doesn't ask you to find a Maclaurin series -- it just asks you to prove the limit that is given.

    The proof seems very straightforward to me. What part of it do you not understand?
     
  4. May 29, 2016 #3
    I don't see how showing that the series converges for all x proves that the limit as n approaches 0 is equal to 0 for all x.
     
  5. May 29, 2016 #4

    jbunniii

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    This is true in general for any convergent series. If ##\sum_{n=0}^{\infty}a_n## converges, then ##\lim_{n \to \infty} a_n = 0##. This is sometimes called the n'th term test when stated in its contrapositive form: if the sequence ##a_n## does not converge to zero, then the series ##\sum_{n=0}^{\infty} a_n## diverges.

    The converse is false, as shown by the example ##\sum_{n=1}^{\infty}(1/n)##, which diverges even though ##\lim_{n\to\infty}1/n = 0##.
     
  6. May 30, 2016 #5
    That's true, but aren't we trying to prove the limit as n approaches 0, not infinity?
     
  7. May 30, 2016 #6

    Mark44

    Staff: Mentor

    What you just wrote makes no sense. They are trying to prove that the limit is zero, as n approaches infinity. More specifically, that ##\lim_{n \to \infty}\frac {x^n}{n!} = 0##. They are using the Ratio Test to prove this.

    In addition, this work also proves that the series ##\sum_{n = 0}^{\infty}\frac{x^n}{n!}## is convergent, for the reason jbunniii gave.
     
  8. May 30, 2016 #7
    The first sentence, they are trying to prove the limit is 0 as n approaches 0.
     
  9. May 30, 2016 #8

    Mark44

    Staff: Mentor

    You're right. The type is so small in the image you posted that I misread the 0 for ∞.

    I'm pretty sure that the ##n \to 0## under "lim" is a typo. The other work shown suggests strongly that it should be ##\lim_{n \to \infty}## which is what I mistook it to be.
     
  10. May 30, 2016 #9
    Ah, I see. How would one go about prove the limit as n approaches 0, or would it be very hard?
     
  11. May 30, 2016 #10

    Mark44

    Staff: Mentor

    No, it's not hard. If ##x \ne 0##, ##\lim_{x \to 0}\frac {x^n}{n!}## = 1. As ##n \to 0, x^n \to 1##, and 0! is defined to be 1. ##0^0## is indeterminate, though, which is why I qualified what I said about x not being equal to zero.
     
  12. May 30, 2016 #11

    jbunniii

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    It doesn't make much sense to talk about the limit as ##n## approaches zero. This is because ##n## is an integer, so the only way it can approach zero is by being zero. So, the limit of a sequence ##a_n## as ##n## approaches zero is simply ##a_0##.
     
  13. Jun 5, 2016 #12
    Perhaps try separating the ##|x|## and the ##\frac{1}{n+1}## and you may see why it works for any real x.

    It has to do with the value of ##\frac{1}{n+1}## as n approaches infinity.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Proving a limit (x^n/(n!))
  1. Prove x^n < n! (Replies: 4)

  2. Prove x^n<y^n (Replies: 1)

Loading...