the sum $\sum \limits_{n>1} f(n)/n$ over primes Let 
$$
f(n)=\begin{cases}-1&\text{if $n$ is a prime integer},\\
1&\text{otherwise}.
\end{cases}
$$ 
Then, does the series
$$
\sum_{n>1} f(n)/n
$$
converge or diverge?
 A: André Nicolas has already shown that the series diverges. The divergence can be quantified with relative ease as well.
$$ S(n) = \sum_{k=1}^{n} \dfrac{1 - 2 \times 1_{\text{$k$ is a prime}}}k = \sum_{k=1}^{n} \dfrac1n - 2 \sum_{k- \text{ prime }\leq n} \dfrac1k \\ = \left(H_n - \log n \right) - 2 \left(\sum_{k- \text{ prime }\leq n} \dfrac1k - \log (\log n) \right) + \log (n) - 2 \log \log (n)$$
From the asymptotic of $H_n$ and $\sum_{k- \text{ prime }\leq n} \dfrac1k$, we have that $$S(n) = \gamma - 2B + \log \left( \dfrac{n}{\log^2 (n)} \right) + \mathcal{O} \left( \dfrac1{\log n}\right)$$
A: Let $f(n)=-1$ if $n$ is prime, and $f(n)=1$ otherwise. We show that the sum 
$$\sum_{n=6}^\infty \frac{f(n)}{n}$$
does not converge.
Arrange the integers $\ge 6$ in consecutive groups of $6$. In the set  $\{6k,6k+1,6k+2,6k+3,6k+4,6k+5\}$, the numbers $6k+1$ and $6k+5$ may be prime. The other four are definitely composite. It follows that
$$\sum_{i=0}^5 \frac{f(6k+i)}{6k+i} \ge \frac{1}{6k}-\frac{1}{6k+1}+\frac{1}{6k+2}+\frac{1}{6k+3}+\frac{1}{6k+4}-\frac{1}{6k+5}.$$
The sum on the right is $\gt \frac{1}{6k+2}+\frac{1}{6k+3}$, which in turn is $\gt \frac{2}{6k+3}$.  
But $\sum_{k=1}^\infty \frac{2}{6k+3}$ diverges.  So the sequence of partial sums of the shape $\sum_{n=6}^{6k+5} \frac{f(n)}{n}$ is unbounded, and therefore the series of the question does not converge.
The argument proves a little more: the series in fact diverges to $\infty$. 
