Sequence of Primes: Concluding Divergence for All e>0

[Dragonfall]

From the fact that \sum_{\mathbb{P}}\frac{1}{p} diverges, how do I conclude that the sequence \frac{n^{1+e}}{p_n} diverges for all e>0?

(p=prime, P_n=nth prime)

 

[shmoe]

What have you tried so far? A hint- try proving this by contradiction.

 

[Dragonfall]

Suppose it converges, then since \sum\frac{1}{n^{1+e}} converges for any e>0, \sum_P\frac{1}{p} must converge as well, which is impossible.

 

[shmoe]

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?