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Sequence of Primes

  1. May 29, 2006 #1
    From the fact that [tex]\sum_{\mathbb{P}}\frac{1}{p}[/tex] diverges, how do I conclude that the sequence [tex]\frac{n^{1+e}}{p_n}[/tex] diverges for all e>0?

    (p=prime, P_n=nth prime)
     
  2. jcsd
  3. May 29, 2006 #2

    shmoe

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    What have you tried so far? A hint- try proving this by contradiction.
     
  4. May 29, 2006 #3
    Suppose it converges, then since [tex]\sum\frac{1}{n^{1+e}}[/tex] converges for any e>0, [tex]\sum_P\frac{1}{p}[/tex] must converge as well, which is impossible.
     
  5. May 30, 2006 #4

    shmoe

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    You might want to give a little more info on why the divergence of the first sum implies the divergence of the second, presumably you're using one of the standard methods of relating the two sums, but which?
     
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