An inequality on the series of powers of reciprocals of the primes 
Let $p_n$ denote the $n$-th prime $(p_1=2)$
Let $s>1$
Prove that $\displaystyle-1+\ln(\frac{s}{s-1})\leq\sum_{k=1}^\infty\frac{1}{p_k^s}\leq\ln(\frac{s}{s-1})$

Using the classical Euler product formula and integrals, I proved that $$\frac{s-1}{s}\leq \prod_{k=1}^\infty(1-\frac{1}{p_k^s})\leq s-1$$
which is enough to get $\displaystyle \sum_{k=1}^\infty\frac{1}{p_k^s}\leq\ln(\frac{s}{s-1})$ (thus $\lim_{s\to \infty}\sum_{k=1}^\infty\frac{1}{p_k^s}=0$)
How can I prove that $\displaystyle-1+\ln(\frac{s}{s-1})\leq\sum_{k=1}^\infty\frac{1}{p_k^s}$ ? The same technique does not apply.

Note that (assuming $s
 $ real) $$\log\left(\frac{s}{s-1}\right)-1>0\Longleftrightarrow s<\frac{1}{e-1}+1
 $$ so for $s\geq1+\frac{1}{e-1}
 $ it's surely true. From the Euler product for Zeta, we have $$\log\left(\zeta\left(s\right)\right)=\sum_{p}\frac{1}{p^{s}}+\sum_{p}\sum_{m\geq2}\frac{1}{mp^{ms}}=\sum_{p}\frac{1}{p^{s}}+C(s)\,\,\,\,\,\,\,(1)
 $$ with $0<C(s)<1
 $. Using $\zeta\left(s\right)=\frac{1}{s-1}+\gamma+o\left(1\right)
 $ at $s\rightarrow1^{+}
 $ we have $$\sum_{p}\frac{1}{p^{s}}=\log\left(\frac{1}{s-1}+\gamma+o\left(1\right)\right)-C(s).
 $$ Note that $(1)
 $ implies, using your bounds, that $$\sum_{p}\frac{1}{p^{s}}\leq\log\left(\frac{s}{s-1}\right)-C(s).
 $$
