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Divergence of 2^(n^2)/n!

  1. Nov 9, 2012 #1
    Hi,
    How may I show that 2^(n^2)/n! converges to infinity?
     
  2. jcsd
  3. Nov 9, 2012 #2

    Dick

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    That is SO divergent you can afford to be pretty sloppy. 2^(n^2)/n!>2^(n^2)/n^n, right? So show 2^(n^2)/n^n diverges. Hint: look at the log.
     
  4. Nov 9, 2012 #3
    Thank you, Dick!
     
  5. Nov 9, 2012 #4
    Would it be correct to say that if for sequences a_n and b_n, lim a_n = infinity and |b_n|< c < infinity, then lim|a_n*b_n| = infinity?
    (I think it should be correct, as we may infer that lim |bn| = c and then the limit of the product of a_n and b_n would yield c*infinity which is always infinity.)
     
  6. Nov 9, 2012 #5

    Dick

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    If you mean lim |b_n|=c with c>0, then sure. If c=0, then you need to think more.
     
    Last edited: Nov 9, 2012
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