= \displaystyle\sum_{n=1}^{\infty} \frac{1}{n^s}$ is absolutely convergent, so we have $$\displaystyle\lim_{s \rightarrow 1^+} \log(\zeta(s)) = \displaystyle\sum_{n=1}^{\infty} \frac{1}{\displaystyle\lim_{s \rightarrow 1^+} n^s} = \displaystyle\sum_{n=1}^{\infty} \frac{1}{n},$$ which is the harmonic series , a well known divergent series. Therefore, we have just shown that $$\infty = \displaystyle\sum_{n=1}^{\infty} \frac{1}{n} = \displaystyle\sum_p \frac{1}{p} + \mathscr{O}(1),$$ confirming that the sum