Proving series convergence I want to show that
$$\prod_{p\in \mathbb{P}}(1-p^{-s})^{-1}$$
converges if Re$(s)&gt;1$, and $\mathbb{P}$ is the set of all prime numbers. In order to show that the infinite product converges, we'll look at 
$$-\sum_{p\in \mathbb{P}}\ln(1-p^{-s}).$$
How should I show that it converges?
Apply some basic results: That $\sum_{n\in N} n^{-t}$ converges for $t&gt;1,$ and these two theorems:

Theorem 1. Let $0\leq a_n&lt;1$ for $n\in N.$ Then $$(1)....\quad \prod_{n\in N}(1-a_n)&gt;0 \iff \sum_{n\in N}a_n&lt;\infty.$$ Theorem 2. Let $b_n\geq 0$ for $n\in N.$ Then $$(2)....\quad \prod_{n\in N}(1+b_n)&lt;\infty \iff \sum_{n\in N}b_n&lt;\infty.$$
 Now for $Re(s)&gt;1$ and prime $p$ we have $$(3)....\quad 0&lt;1-p^{-Re(s)}\leq |1-p^{-s}|\leq 1+p^{-Re(s)}.$$ Therefore  $$(4)....\quad \prod_{p\in P}(1-p^{-Re(s)})\leq   \prod_{p\in P}|1-p^{-s}|\leq \prod_{p\in P}(1+p^{-Re (s)}).$$ By Theorem 1  we have $$(5)....\quad 0&lt;\prod_{n\in N}(1-n^{-Re(s)})\leq \prod_{p\in P}(1-p^{-Re(s)}).$$ By Theorem 2 we have $$(6)....\quad \infty&gt;\prod_{n\in N}(1+n^{-Re(s)})\geq \prod_{p\in P}(1+p^{-Re(s)}).$$ So $$(7)....\quad 0&lt;\prod_{p\in P}|1-p^{-s}|&lt;\infty.$$ So, with $s_n=\prod_{p\in P \land p\leq (n+1)}(1-p^{-s} )$, the sequence $S=(s_n)_{n\in N}$  is bounded in modulus by a positive lower bound and a finite upper bound. Therefore $S$ has a convergent subsequence. We can now use the  inequalities of (3), the convergence of the products in (4) &amp; (5), and the basic $\epsilon, \delta$ def'n of convergence to show then that $S$ has a unique non-zero limit.
As it relates to the Reimann zeta function.

Euler showed that:

$\zeta (s) = \sum_\limits{n=1}^{\infty} n^{-s} = \prod_{p\in \mathbb{P}}(1-p^{-s})^{-1}$

https://en.wikipedia.org/wiki/Proof_of_the_Euler_product_formula_for_the_Riemann_zeta_function

Now we just have to prove that $\sum_\limits{n=1}^{\infty} n^{-s}$ converges when $s&gt;1.$

The most direct way would be the integral test. 

Following the logic above:

$-\sum_{p\in \mathbb{P}}\ln(1-p^{-s})$

$p^{-s}&lt;-\ln (1-p^{-s})&lt;(p^s-1)^{-1}$

and the $n{th}$ prime number is always greater than $n.$

If $\sum_{n\in \mathbb{N}}n^{-s}$ converges, $-\sum_{p\in \mathbb{P}}\ln(1-p^{-s})$ converges by the comparison test.