Sum f(p) over all primes convergent with sum f(n) over all natural numbers divergent?

The sum $\sum_{n=1}^{\infty} 1/n^{s}$ is convergent for all real $s>1$ and diverges for all real $s \le 1$. The same holds for the sum $\sum_{p \ prime} 1/p^{s}$. Thus, for the functions $f(n)= 1/n^s, s \in \mathbb{R}$ the sum $\sum_{n=1}^{\infty}f(n)$ shows the same convergence behaviour as the sum $\sum_{p \ prime}f(p)$.

The same holds, if I'm not mistaken, for the functions $f(n)= 1/(n (\ln n)^s), s \in \mathbb{R}$ (both for $n \in \mathbb{N}$ and for primes convergence iff $s>1$).

Question: Is there a real monotonic function $f$ such that $\sum_{n=1}^{\infty}f(n)$ diverges whereas sum $\sum_{p \ prime}f(p)$ converges? (The monotony requirement is for preventing 'artificial' solutions that single out the primes (as e.g. $f(n) = 2^n$ if $n$ is prime; $f(n)=n$ if $n$ is not prime)).

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The simple approach to this problem is to replace \sum_p f(p) with the roughly equivalent \sum_n f(n log n) since the nth prime is roughly n log n. Once you've found a monotonic function where \sum_n f(n log n) converges but \sum_n f(n) diverges then it probably won't be too hard to use the prime number theorem to answer your question. – Noah Snyder Jul 6 '10 at 21:12
Or, easier to think about, a function $f(t)$ such that $\int f(t) dt$ converges and $\int f(t) dt/\log t$ diverges. – David Speyer Jul 6 '10 at 21:15
f(n) = n log n has that property, doesn't it? – Qiaochu Yuan Jul 6 '10 at 21:22
Switch converges and diverges in my comment. – David Speyer Jul 6 '10 at 21:24
Qiaochu, I think you mean $f(n) = 1/(n \log n)$. – Michael Lugo Jul 6 '10 at 22:05

I think, you are mistaken, sum $\sum 1/(p\log p)$ converges, since $p_n\log p_n$ behaves like $n(\log n)^2$
Note also that if $\{a_n\}$ is an increasing sequence of natural numbers that presents arbitrarily large gaps between consecutive terms (e.g. a sequence with with density $0$), there is always a positive decreasing function $f$ such that the sum of $f(a_n)$ converges and the sum of $f(n)$ diverges to positive infinity. The reason is that $\sum_n f(n)\geq\sum_n f(a_n)(a_n-a_{n-1}),$ and as the $a_n-a_{n-1}$ are unbounded, the claim reduces to the easy: for any unbounded sequence $u_n>0$ there is a convergent series with coefficients $\epsilon_n>0$ such that the series of the $u_n\epsilon_n$ diverges. Use this taking $u_n:=a_n-a_{n-1}$ to define a decreasing $f$ such that $f(a_n)=\epsilon_n$.