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Homework Help: 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


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    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


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    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]
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