Difficulty with a meromorphic extension. I'm trying to understand the prime number theorem, but never having followed a course in complex analysis, I have some difficulties. (the article is this: http://www.maths.dur.ac.uk/~dma0hg/prime_number_theorem_zagier.pdf)
I understand intuitively what an holomorphic extension is, but some implications are obscures for me.
In particular when it's proved that $-\frac{\zeta'(s)}{\zeta(s)} = \sum_{p} \frac{log(p)}{(p^s-1)} =\phi(s) + \sum_p \frac{log(p)}{p^s(p^s-1)} $
it's directly stated that the final sum converges for $\Re(s)>1/2$ (Why?)
moreover this, and the fact that $\zeta(s) - \frac{1}{s-1}$ is holomorpic for $\Re(s)>0$ implies that $\phi(s)$ extends meromorphically to $\Re(s)>1/2$.
Why this implication?
Thank you for the help!
 A: The quantity $\log n/n^s(n^s - 1)$ is comparable to $\log n/n^{2s}.$
Now the series $\zeta(s) = \sum_n 1/n^s$ converges (absolutely, and uniformly on compact subsets) when $\Re s > 1$, so the same is true of its derivative,
which is $-\sum_n \log n/n^s$.  Thus $\sum_n \log n /n^{2 s}$ converges if $\Re s > 1/2,$ and so the same is true for $\sum_{n = 2}^{\infty} \log n/ n^s (n^s -1)$,
and hence the corresponding some over primes also converges (since it is bounded above by the sum over all $n \geq 2$).  In fact, the convergence is absolute (again, by comparison with the $\zeta$ function), and uniform on compact subsets, and hence the limit is a holomorphic function of $s$ (provided $\Re s > 1/2$).
Since $\zeta(s) - 1/(s-1)$ is holomorphic when $\Re s > 0$, we see
that $\zeta(s)$ is meromorphic in this region (being the sum of a holomorphic function and the meromorphic function $1/(s-1)$).  Thus $\zeta'(s)$ is also
meromorphic in this region, and hence so is the ratio $\zeta'(s)/\zeta(s)$.
Finally, $\phi(s)$ is the difference of a meromorphic function and a holomorphic function on the region $\Re s > 1/2$, and so is meromorphic in this region.
