$\zeta (s+c)/\zeta (s)$ is increasing It appears that for the Riemann zeta function
$$\zeta (s) = \sum_{n\ge 1} \frac{1}{n^s} \quad (s > 1)$$
and any constant $c > 0$, the function
$$\frac{\zeta (s+c)}{\zeta (s)}$$
(considered for real $s > 1$) is increasing.
Could anybody give a proof of this fact?
Thank you very much!

Just in case, here's what I tried: we have
$$\left(\frac{\zeta (s+c)}{\zeta (s)}\right)' = \frac{\zeta (s) \zeta' (s+c) - \zeta (s+c) \zeta'(s)}{\zeta (s)^2},$$
and $\zeta' (s) = -\sum_{n\ge 2} \frac{\log n}{n^s}$, but after playing around with series I don't see how to deduce that $\zeta (s) \zeta' (s+c) > \zeta (s+c) \zeta'(s)$.

Since $s>1$, $\zeta(s)>0$ and it's easier to consider the logarithmic derivative: $f'(x)/f(x)$ has the same sign as $f'(x)$. In particular, for $s>1$ we can use the Euler product 
$$ \zeta(s) = \prod_{p} (1-p^{-s})^{-1} : $$
from this we have
$$ \frac{d}{ds} \log{\zeta(s)} = \frac{d}{ds} \sum_p -\log{(1-p^{-s})} = -\sum_p \frac{1}{p^s-1} \log{p} . $$
Then
$$ \frac{d}{ds} \log{\frac{\zeta(s+c)}{\zeta(s)}} = \frac{d}{ds} \log{\zeta(s+c)} - \frac{d}{ds} \log{\zeta(s)} = \sum_p \left( \frac{1}{p^{s}-1} - \frac{1}{p^{s+c}-1} \right) \log{p} . $$
Since $s \mapsto 1/(p^s-1)$ is a decreasing function, every term in the sum is positive, so the whole sum is positive. Therefore $s \mapsto \zeta(s+c)/\zeta(s)$ is in fact strictly increasing when $c>0$ and $s>1$.
(All the interchanges of logarithms and derivatives with sums and products can be justified using continuity and local uniform convergence, as usual: there's nothing weird going on here.)
