i am stuck on proving this property of a series:

let p_n=sigma (from k=1 to n) (1/k) and p=sigma (from k=1 to infinite) (1/k).

Prove for all N, a natural number, there exists a n_0 in the naturals such that abs (p_n-p_n-1)<=1/N for all n>=n_0 yet abs (p-p_n)> 1/N for all n>=n_0.

essentially one would need to show that abs (1/n) <= 1/N and that sigma(from k=n+1 to infinite) (1/n)>=1/N. how would we choose our n_0 in a way that this would work? thanks!