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Homework Help: Where am I going wrong

  1. Sep 28, 2007 #1
    1. The problem statement, all variables and given/known data

    I stumbled at the answer of a problem solved in Reily-Hobson-Bence.Please check if I am wrong:

    Find the parts of the z-plane for which the following series is convergent:

    ∑[(1/n!)(z^n)] where n runs from 0 to ∞

    2. Relevant equations
    3. The attempt at a solution

    Cauchy's radius: (1/R)=Lt (n->∞) [(1/n!)^(1/n)]

    Changing variables as 1/n=t,the limit becomes (1/R)=Lt(t->0) [(t!)^t]
    I am having the limit as 1 and hence,R=1...but the book says Since
    [(n!)^(1/n)] behaves like n as n → ∞ we find lim[(1/n!)^(1/n)] = 0. Hence R = ∞ and the series is convergent for all z
  2. jcsd
  3. Sep 28, 2007 #2
    Why do you think you're wrong?
  4. Sep 28, 2007 #3
    I do not really think IO am wrong...But I am not the author,you know...might be I am missing something.
    Also,in the very next problem [with (1/n!) in this problem replaced by (n!)] the author uses the same principle to say the limit is infinity...
  5. Sep 28, 2007 #4


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    Which one is larger:
  6. Sep 28, 2007 #5


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    It really doesn't make a whole lot of sense to say that (t!)t goes to 1 as t goes to 0. How are you defining t! for t< 1?
  7. Sep 28, 2007 #6
    yes,i was wrong there...
  8. Sep 28, 2007 #7
    NateTG,I get your point...I think there is no other calculus method other than this observation rule...(I was refering to some limit method...)Because,the function is discrete,possibly no standard calculus will evaluate this...

    I may frame the logic like this:

    n! increases more rapidly than n
    =>(1/n!) decreases more rapidly than (1/n)
    here n is integer...for n--->∞, (1/n!)--->0 (also<1) and its rate of fall must be more than that of

    Clearly, the limit--->0
  9. Sep 28, 2007 #8
  10. Sep 29, 2007 #9


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    I would be careful with stirling's approximation. It's a nice magic bullet, but it may not help you understand things.
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