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!

Convergence of 1/K!

  1. Apr 15, 2015 #1
    1. The problem statement, all variables and given/known data

    Does (1/(k!)) converge?

    2. Relevant equations

    Convergence Tests?


    3. The attempt at a solution

    I thought I could just simply use the divergence test, but I'm not sure if that only tells you if it's divergent and not whether it is convergent or not.

    lim(k>inf) (1/(k!)) = 0, if L =/= 0, then it is divergent, but since it is 0 does that make it convergent, or inconclusive?

    I've thought about trying a comparison test but I'm not sure what I could compare 1/k! too, would I compare it to 1/k? In which case it would be divergent.

    Kinda stuck here on which to use.
     
  2. jcsd
  3. Apr 15, 2015 #2
    Are you wondering about the convergence of the series ##\Sigma_{k=0}^\infty \frac{1}{k!}##

    We can set ##e^x=\Sigma_{k=0}^\infty \frac{x^k}{k!}##. Then ##e^1=\Sigma_{k=0}^\infty \frac{1^k}{k!}=\Sigma_{k=0}^\infty \frac{1}{k!}##

    So, the series converges and equals ##e##.
     
  4. Apr 15, 2015 #3

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    While it is true that [itex]\sum \frac{1}{k!}[/itex] converges to e, if you did not know that initially, you could have used the "ratio test": [itex]\sum a_n[/itex] converges if [itex]\frac{a_{n+1}}{a_n}< 1[/itex] and diverges if it is greater than one.

    (Your test, that if [itex]\sum a_n[/itex] converges, the [itex]a_n[/itex], must go to 0, is generally called the "divergence test" since it tells us that if that limit is NOT 0 then the series diverges but does NOT tell us one way or the other if the limit is 0.)
     
  5. Apr 15, 2015 #4

    phyzguy

    User Avatar
    Science Advisor

    I don't think you stated the ratio test quite right. The ratio test tells us that the series converges if [itex]lim_{n->\infty} \frac{a_{n+1}}{a_n} <1[/itex]. The series the OP asked about certainly passes this test, since [itex]lim_{n->\infty} \frac{n!}{(n+1)!} = lim_{n->\infty} \frac{1}{n+1} = 0[/itex]. You wrote that the ratio test is that the series converges if [itex]\frac{a_{n+1}}{a_n} < 1[/itex]. The series [itex]\sum_{n=1}^\infty \frac{1}{n} [/itex] would pass this test since [itex]\frac{a_{n+1}}{a_n} = \frac{n}{n+1} < 1[/itex] for all n, but it does not converge. The ratio test for this series gives [itex]lim_{n->\infty} \frac{a_{n+1}}{a_n} = lim_{n->\infty} \frac{n}{n+1} = 1[/itex]
     
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: Convergence of 1/K!
Loading...