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Quick convergent series question.

  1. Dec 30, 2012 #1
    2^n + n!/4^n ?
    so by the 'vanishing condition' as n!/4^n does not ---> 0 as n --> infinity, this part of the series diverges.
    however (e.g via the ratio test 2^n/4^n converges).
    My book concludes that due to part a, the entire series divegres. However I am struggling to see how this justifies that the whole series will diverge - I thought that if part of a series diverges and part diverges , than what the whole series does depends upon the exact convergence and divergence - i.e could diverge or converge?
     
  2. jcsd
  3. Dec 30, 2012 #2

    lurflurf

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    The whole series will diverge by comparison

    n!/4^n<(2^n + n!)/4^n

    If the parts were of opposite sign there would be the possibility of convergence.
     
  4. Dec 30, 2012 #3

    Ray Vickson

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    Looking at what you wrote, which is
    [tex] 2^n + \frac{n!}{4^n},[/tex]
    we have, of course, that both parts diverge. However, assuming you meant
    [tex] \frac{2^n + n!}{4^n}, [/tex]
    which, for some reason you did not want to write as (2^n + n!)/4^n, then you are correct: one "part" converges and the other diverges. However, that *automatically* means that the total diverges. Suppose we have two parts are ##t_1(n)## and ##t_2(n)## and that##\sum t_1(n)## converges while ##\sum t_2(n)## diverges. We have, for finite N:
    [tex] S(N) = \sum_{n=1}^N (t_1(n) + t_2(n)) = S_1(N) + S_2(N), [/tex] where the ##S_i(N)## are the partial sums, and our assumptions are that
    [tex] \lim_{N \to \infty} S_1(N) = s_1 [/tex] exists and is finite, while
    [tex] \lim_{N \to \infty} S_2(N) [/tex] does not exist (that is, it is either ##\pm \infty## or else does not exist at all, due to oscillations, etc). These two statements imply that S(N) does not have a finite limit as N → ∞, so the total series diverges.
     
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