# Euler Product formula for Riemann zeta function proof

In class we introduced Reimann Zeta function

$$\zeta (x)=\sum_{n=1}^{+\infty} \frac{1}{n^x}$$ And we proved its domain was $D=(1,+\infty)$

Now Euler proved that

$$\zeta(x)=\prod_{p\text{ prime}}\frac{1}{1-p^{-x}}$$

By saying $$\zeta(x)=1+\frac{1}{2^x}+\frac{1}{3^x}+... \\ \zeta(x)(\frac{1}{2^x})=\frac{1}{2^x}+\frac{1}{4^x}+... \\ \zeta(x)(1-\frac{1}{2^x})=1+\frac{1}{3^x}+\frac{1}{5^x}+...$$

And so on for every prime number.

However this proof isn't a 'rigorous proof' as my professor says. Why is that and how would one prove this rigorously? Any reference would be helpful. I have seen on wikipedia that to make the proof rigorous we need to observe $\mathfrak{R}(x)>1$ Is that the real part of x or something else?

• it is rigorous if you prove first that both side converge, and that you continue your sieve, finding an induction argument on the $k$th prime. and yes the Euler product is true for any $x \in \mathbb{C}$ such that $Re(x) > 1$, but considering only $x$ real and $> 1$ is what Euler did. Commented Apr 14, 2016 at 21:14
• If you want to make it rigorous, first make sure you know what an infinite product means... Commented Apr 14, 2016 at 21:16
• I guess i do know what infinite product means @barto :) Why would you think otherwise ? Commented Apr 14, 2016 at 21:17
• And regarding the equation the professor wasn't clear that we need to prove it for $x\in \mathbb{C}$ so im going to guess $x\in \mathbb{R}$ and $x>1$. And i'm not quite sure i follow @user1952009 does the induction go by prime numbers ? Do i prove the convergence on both sides for the induction step and basis ? Commented Apr 14, 2016 at 21:20
• The book "Gamma: Exploring Euler's Constant by Julian Havil" -- I am pretty sure that you will read not only the easiest but also the most rigorous proof of the theorem.
– user231343
Commented Apr 14, 2016 at 21:57

(this is how I'd do it)

consider the formal product and series, then by induction on the $k$th prime : $$\prod_p (1+p^{-x}+p^{-2x}+\ldots) = \sum_n a_n n^{-x}$$

now consider the coefficient $a_1$ : it is clearly $1$, the coefficient $a_2$ : it is clearly $1$, etc. (by the fundamental theorem of arithmetic).

now do the same with $$F_K(x) = \prod_{p \le K} (1+p^{-x}+p^{-2x}+\ldots) = \sum_{n=1}^\infty a_n(K) n^{-x}$$

then if $n = \prod_i p_i^{e_i}$ : $a_n(K) = 1$ if all the $p_i \le K$, otherwise $a_n(K) = 0$.

clearly $F_K(x)$ is well-defined for any $x > 1$ (it is a finite product), and $\lim_{K \to \infty} F_K(x)$ exists too because the logarithm of the infinite product is $-\sum_p \ln(1-p^{-x})$ which is absolutely convergent since $\ln(1-p^{-x}) \sim -p^{-x}$ and that $\sum_p p^{-x} < \sum_{n=1}^\infty n^{-x}$ which is (absolutely) convergent.

finally, $\zeta(x)- F_K(x) = \sum_{n=1}^\infty |a_n(K)-1| n^{-x} > 0$, it is absolutely convergent, it is decreasing in $K$, and it clearly $\to 0$ when $K \to \infty$ since every term $\to 0$.

i.e. :

$$\lim_{K \to \infty} F_K(x) = \prod_p \frac{1}{1-p^{-x}} = \zeta(x) \qquad\qquad (\forall \ x > 1)$$

the proof for every $Re(x) > 1$ is a little more complicated, since we don't have monotone convergence of $\zeta(x)-F_K(x)$ to $0$ but only absolute convergence.

• How do you know that $\prod_{p} \sum_{n} p^{-x}$ can be written as $\sum_{n} \frac{a_n}{n^x}$ and how do you see that those relations for $a_n$ apply since a unique decomposition of every natural number by primes don't seem to imply that (or at least not that i see) Commented Apr 15, 2016 at 10:40
• @daniels_pa : I wrote it, by induction. but it was just to show that formally the equality was true. then I showed how to prove rigorously that the functions of $x$ are the same. and if you don't see how the fundamental theorem of arithmetic applies to $\prod_p (\sum_{\nu = 0}^\infty p^{- \nu x})$ it is not my bad. Commented Apr 15, 2016 at 15:48
• in one word : work more. Commented Apr 15, 2016 at 15:50
• I understand what you wanted but the induction goes like for 2 : $\sum_{n=1}^{+\infty} \frac{a_n}{n^x} = \sum_{n=1}^{+\infty} \frac{1}{2^{nx}}$ that said $a_n$ will be 1 for all $2^k$ where $k\in\mathbb{N}$ From there the induction step was bit problematic for me to get but i did finally understood it. I understand your proof that the functions of x are the same from the start. There is no need to be rude :) Sometimes we are all tired and short with time. Commented Apr 15, 2016 at 19:43
• @daniels_pa : finally, the simplest way you should remember is to say that : $\lim_{N \to \infty} \sum_{n \le N}$ and $\lim_{K \to \infty} \prod_{p \le K}$ are just to different orders for summing $(n^{-x})_{n \in \mathbb{N}^*}$, and since for $Re(x) > 1$ it is an ABSOLUTELY convergent series, hence the order of summation doesn't matter, and the two order of summation give the same result (and hence the same function of $x$ : $\zeta(x)$) Commented Apr 15, 2016 at 20:00

Eulers formula for the Zeta function is, $$\prod_{p \in \mathbb{P}}^{p \le A} \frac1{1- p^{-s}} = \prod_{p \in \mathbb{P}} (\sum_{k=0}^{\infty} p^{-ks})$$

Infinite sums and products may depend on order. Finite do not. Consider the finite products, $$\prod_{p \in \mathbb{P}}^{p \le A} \frac{1- p^{-(K+1)s}}{1- p^{-s}} = \prod_{p \in \mathbb{P}}^{p \le A} \sum_{k=0}^{K} p^{-ks}$$

Product over a sum is the sum over the cartesian products of the products. $$\prod_{a \in A} \sum_{b \in B_a} t_{a,b} = \sum_{c \in \prod_{a \in A} B_a} \prod_{a \in A}t_{a,c_a}$$

where the product of sets $\prod_{a \in A} B_a$ is taken to be a cartesian product.

$$\prod_{p \in \mathbb{P}}^{p \le A} \sum_{k=0}^{K} p^{-ks} = \sum_{v \in \prod_{p \in \mathbb{P}}^{p \le A} \{ 0 .. K\}} \prod_{p \in P}^{p \le A}p^{-v_ps} = \sum_{v \in \prod_{p \in \mathbb{P}}^{p \le A} \{ 0 .. K\}} (\prod_{p \in P}^{p \le A}p^{-v_p})^{-s}$$

Change the summing variable using, $$\sum_{v \in V } g(f(v)) = \sum_{w \in \{f(v) : v \in V \} } g(w)$$ which is valid only if f is one to one. This is true by Fundamental theorem of arithmetic, as every number has a unique factorization. This gives,

$$\prod_{k=0}^{K} \sum_{p \in \mathbb{P}}^{p \le A} p^{-ks} = \sum_{n \in \{\prod_{p \in P}^{p \le A}p^{-v_p} : v \in \prod_{p \in \mathbb{P}}^{p \le A} \{ 0 .. K\} \} } n^{-s}$$

Define Q by, $$Q(A, K) = \{\prod_{p \in P}^{p \le A}p^{-v_p} : v \in \prod_{p \in \mathbb{P}}^{p \le A} \{ 0 .. K\} \}$$

No natural number may have a factor greater than itself. Also the highest power it can have for a prime factor is $N = 2^K$, giving $K = \log_2(N)$. Every number from 1..N must have a unique factorization, and that factorization must be constructed by Q.

$$\{1 .. N\} \subset Q(N, log_2(N))$$

The positive natural numbers are given by, $$\lim_{A \to \infty, K \to \infty}Q(A,K) = \mathbb{N+} = \{\prod_{p \in P}p^{-v_p} : v \in \prod_{p \in \mathbb{P}} \{ 0 .. \infty\} \}$$

Then, $$\prod_{p \in \mathbb{P}}^{p \le A} \frac{1- p^{-(K+1)s}}{1- p^{-s}} = \prod_{p \in \mathbb{P}}^{p \le A} (\sum_{k=0}^{K} p^{-ks}) = \sum_{n \in Q(A, K)} n^{-s}$$

Taking limits, $$\lim_{A \to \infty, K \to \infty} \prod_{p \in \mathbb{P}}^{p \le A} \frac{1- p^{-(K+1)s}}{1- p^{-s}} = \prod_{p \in \mathbb{P}} \frac1{1- p^{-s}}$$

$$\lim_{A \to \infty, K \to \infty} \sum_{n \in Q(A, K)} n^{-s} = \sum_{n \in \mathbb{N^+}} n^{-s}$$

The order of summation is not prescribed by the sum. Infinite sums can have different values depending on order. However, $$\sum_{1..n}^{\infty} n^{-s}$$ is absolutely convergent for $\Re(s) > 1$, which guarantees that the sum will converge to the same limit irrespective of order. So $$\prod_{p \in \mathbb{P}} \frac1{1- p^{-s}} = \sum_{n \in \mathbb{N^+}} n^{-s} = \sum_{n = 1}^{\infty} n^{-s} = \zeta(s)$$

An alternative approach compares the limit with, $\zeta(s)$. Consider, $$\prod_{p \in \mathbb{P}} \frac1{1- p^{-s}}- \zeta(s)$$

Then, $$\lim_{A \to \infty, K \to \infty} \prod_{p \in \mathbb{P}}^{p \le A} \frac{1- p^{-(K+1)s}}{1- p^{-s}} - \lim_{N \to \infty} \sum_{n = 1}^{N} n^{-s}$$

Or, $$\lim_{A \to \infty, K \to \infty} \sum_{n \in Q(A, K)} n^{-s} - \lim_{N \to \infty} \sum_{n = 1}^{N} n^{-s}$$

So, $$\lim_{N \to \infty} \sum_{n \in (Q(N, \log_2(N)) - \{1 .. N\})} n^{-s}$$

$s$ may be complex, $s = u + it$ $$\lim_{N \to \infty} \sum_{n \in (Q(N, \log_2(N)) - \{1 .. N\})} n^{-u} e^{-it \ln(n)}$$

The magnitude is,

$$\lim_{N \to \infty} | \sum_{n \in (Q(N, \log_2(N)) - \{1 .. N\})} n^{-s} | <= \lim_{N \to \infty} \sum_{n \in (Q(N, \log_2(N)) - \{1 .. N\})} n^{-u}$$ $$<= \lim_{N \to \infty}\sum_{n=N+1}^{\infty} n^{-u} = 0$$

As $\sum_{n=0}^{\infty} n^{-u}$ converges absolutely for $u > 1$

So if $s = u + it \wedge u > 1$, $$\prod_{p \in \mathbb{P}} \frac1{1- p^{-s}} = \zeta(s)$$

• Hi, you meant $\displaystyle\prod_{p \in \mathbb{P}} (\sum_{k=0}^\infty p^{-ks})$. For the convergence, the main step is $\displaystyle\zeta(s)-\prod_{p \in \mathbb{P},p \le A} (\sum_{k \ge 0} p^{-ks}) = \sum_{n=1}^\infty n^{-s}1_{\text{Lpf}(n) >A}$ where $\text{Lpf}(n)$ is the largest prime factor Commented Sep 7, 2017 at 1:11
• Isn't that bracketing implied? Sorry I don't understand that convergence argument. We are rearranging the terms on an infinite sum. But the euler product converges. So this rearrangement should not matter. Why do you need this extra step? Please explain. Commented Sep 7, 2017 at 1:23
• For $\Re(s) > A$ : $\displaystyle\zeta(s)-\prod_{p \in \mathbb{P}} \frac{1}{1-p^{-s}} = \lim_{A \to \infty} \displaystyle\zeta(s)-\prod_{p \in \mathbb{P},p \le A} \sum_{k \ge 0} p^{-ks} = \lim_{A \to \infty} \sum_{n=1}^\infty n^{-s}1_{\text{Lpf}(n) >A} = 0$. Equivalently, since $\{ n^{-s}\}$ is absolutely summable for $\Re(s)> 1$ then $\lim_{N \to \infty} \sum_{n\le N} n^{-s} = \lim_{A \to \infty} \sum_{\text{Lpf}(n) \le A} n^{-s}$ Commented Sep 7, 2017 at 1:27
• Yes I see that, but why do we need it, seeing as the euler product converges $s > 1$. I thought that if a series converged, you could perform the summation in any order, without changing the value. Just interested to know. Commented Sep 7, 2017 at 1:41
• (if a series converges absolutely then the order of summation doesn't matter) This is exactly what I did in the last formula. Commented Sep 7, 2017 at 1:46

The condition $$\mathrm{Re}(s) > 1$$, where $$\mathrm{Re}(s)$$ is real part of $$s \in \mathbb{C}$$ ensures the $$\zeta$$ series is absolutely convergent and that the infinite product in Euler's product formula converges, and that $$\left| 1 / n^s \right| < 1 / n < 1$$ for all $$n \geq 2$$. The condition comes from the fact that if $$s = \sigma + it \in \mathbb{C}$$, where $$\sigma$$ and $$t$$ are the real and imaginary parts of $$s$$, then $$\left| 1 / n^s \right| = \left| 1 / n^{\sigma} \right|$$, which is the 'p-series' with $$p = \sigma$$, which converges for $$p > 1$$ and diverges for $$p \leq 1$$.

A formal proof based on the sieving method described in Proof of the Euler product formula for the Riemann zeta function is given below, along with a 'heuristic' proof - ie a non-rigorous argument that shows the result might be true. The set of all primes is denoted as $$p_1 < p_2 < \cdots$$. The exponent $$n^s$$ is the principal branch of the complex exponent function $$\exp (s\,\mathrm{Log} \,n)$$ where $$\mathrm{Log}$$ is the principle branch of the complex log function, $$\mathrm{Log}\,z = \log |z| + i\mathrm{Arg}\,z, \forall z \in \mathbb{C} \setminus \{0\}$$ and $$\mathrm{Arg}$$ is the principle complex argument function. All the exponent laws that we assume below are thus valid. (Note standard exponent laws don't generally hold for complex exponent, even on the principle branch - eg $$(e^{z_1})^{z_2} = e^{z_1 z_2}$$ does not in general hold).

## Heuristic Proof

Since $$\left| 1 / p^s \right| < 1$$ the geometric series expansion gives : $$\begin{eqnarray} \prod_p \frac{1}{1 - p^{-s}} & = & \prod_p \left( \sum_{k = 0}^\infty (p^{-s})^k \right) \nonumber \\ & = & \prod_p \left( \sum_{k = 0}^\infty (p^k)^{-s} \right) \hspace{2em} \mbox{(using exponent law)} \nonumber \\ & = & \prod_p \left( 1 + p^{-s} + (p^2)^{-s} + \cdots \right) \nonumber \\ & = & \left( 1 + p_1^{-s} + (p_1^2)^{-s} + \cdots \right) \left( 1 + p_2^{-s} + (p_2^2)^{-s} + \cdots \right) \cdots \label{eq:inf-prod-of-inf-sums} \tag{1} \end{eqnarray}$$

ie. we have an infinite product of infinite sums. Imagine we could treat that in the same way as a finite product of finite sums and multiply out the brackets by summing all products $$q$$ of terms taken one from each bracket in every possible combination (as described at beginning of this answer to Alternative proofs of Euclid-Euler theorem. Then every such product $$q$$ comprises either finitely or infinitely many $$p_r$$ terms. In the latter case each term in the product $$q$$ has form $$1 / (p_r^k)^s$$ where $$k \geq 1$$. But $$\left| 1 / (p_r^k)^s \right| < 1 / p_r^k \leq \frac{1}{2}$$, therefore intuitively, as infinitely many such terms are being multiplied, $$|q| \leq \left( \frac{1}{2} \right)^{\infty} = 0$$. Thus the overall contribution to the expansion's summation from such products is zero.

Consider now a product $$q$$ where only finitely many $$p_r$$'s are involved, ie : $$q = \left( q_1^{a_1} \right)^{-s} \cdots \left( q_m^{a_m} \right)^{-s}$$

for some $$m \in \mathbb{W}$$ (ie integers $$\geq 0$$), certain distinct primes $$q_1, \ldots, q_m$$, and $$a_r \geq 1 \: \forall r \in [1, m]$$.

Then : $$$$q = \left( q_1^{a_1} \cdots q_m^{a_m} \right)^{-s} = n^{-s} = \frac{1}{n^s} \hspace{2em} \mbox{(using exponent law)} \label{eq:finite-term} \tag{2}$$$$

where $$n \in \mathbb{N}$$. All possible combinations of distinct primes $$q_1, \cdots, q_m$$ arise in (\ref{eq:finite-term}) due to the infinite product in (\ref{eq:inf-prod-of-inf-sums}), and all possible powers $$a_r \geq 1 \: \forall r \in [1, m]$$ of those primes arise due to the infinite sums in (\ref{eq:inf-prod-of-inf-sums}). And each combination of these prime powers appears exactly once on expanding the brackets in (\ref{eq:inf-prod-of-inf-sums}). Thus every unique $$n \in \mathbb{N}$$ is produced exactly once, by the Fundamental Theorem of Arithmetic. (Note $$n = 1$$ arises as the product of 'no primes' ie the empty product - by convention equal to $$1$$).

Thus the overall contribution to the expansion's summation of the products $$q$$ containing only finitely many $$p_r$$'s is the sum of (\ref{eq:finite-term}) over all $$n \in \mathbb{N}$$, ie it is the original Zeta function summation (the order of summation of which does not matter as it is absolutely convergent). Thus (\ref{eq:inf-prod-of-inf-sums}) equals the Zeta function, as required.

## Formal Proof

We require to show $$\zeta(s) = \displaystyle{\lim_{k \rightarrow \infty}} a_k \neq 0$$, where :

$$a_k = \left( \frac{1}{1 - p_1^{-s}} \right) \left( \frac{1}{1 - p_2^{-s}} \right) \cdots \left( \frac{1}{1 - p_k^{-s}} \right)$$

Note no denominator is zero since $$\mathrm{Re}(s) > 1$$, and that for any prime $$p$$, $$\left| 1 / p^s \right| < 1 / p < 1$$. Thus it suffices to show $$\frac{1}{a_k} \zeta(s) \rightarrow 1$$ for then $$\zeta(s) \neq 0$$ and $$\frac{1}{a_k} \rightarrow \zeta(s)^{-1} \therefore a_k \rightarrow \zeta(s)$$.

Below we make use of the fact that for any infinite series in $$\mathbb{C}$$ we can arbitrarily insert or remove finitely many zeros between the terms in any manner and it does not alter the sum of the series. We use the term 'sub-series' of an infinite series to mean a series formed from a sub-sequence of the terms of the original series.

We require to show : $$\frac{1}{a_k} \zeta(s) = \left( 1 - \frac{1}{p_k^s} \right) \cdots \left( 1 - \frac{1}{p_1^s} \right) \zeta(s)$$ converges to $$1$$. Firstly, multiplying by the first term : $$\begin{eqnarray*} \left( 1 - \frac{1}{p_1^s} \right) \zeta(s) & = & \sum_{n = 1}^{\infty} \frac{1}{n^s} \; - \; \sum_{n = 1}^{\infty} \frac{1}{(p_1 n)^s} \\ & = & S_1 - S_1', \hspace{2em} \mbox{say} \\ \end{eqnarray*}$$

where $$S_1'$$ is the sub-series of $$S_1$$ comprising all indices divisible by $$p_1$$. Insert zeros into $$S_1'$$ so it 'lines up' with $$S_1$$, then subtract $$S_1'$$ term by term from $$S_1$$. This leaves the original $$S_1$$ with all terms with $$p_1 \mid n$$ 'punched out' to zero. Now we can 'collapse' these punched out zero terms from the series to be left with : $$$$\left( 1 - \frac{1}{p_1^s} \right) \zeta(s) = S_1 - S_1' = \sum_{p_1 \nmid n} \frac{1}{n^s} \label{eq:term-1} \tag{3}$$$$

ie. the original $$\zeta(s)$$ series with all $$n \in \mathbb{N}$$ with $$p_1 \mid n$$ 'sieved out'. Next on multiplying by the $$p_2$$ factor we have : $$\begin{eqnarray*} \left( 1 - \frac{1}{p_2^s} \right) \left( 1 - \frac{1}{p_1^s} \right) \zeta(s) & = & \sum_{p_1 \nmid n} \frac{1}{n^s} \; - \; \sum_{p_1 \nmid n} \frac{1}{(p_2 n)^s} \\ & = & S_2 - S_2', \hspace{2em} \mbox{say} \\ \end{eqnarray*}$$

where $$S_2'$$ is a sub-series of $$S_2$$, since any $$p_2n$$ with $$p_1 \nmid n$$ has $$p_1 \nmid p_2n$$. Thus any term $$p_2n$$ in $$S_2'$$ has $$p_1 \nmid p_2n$$ but $$p_2 \mid p_2n$$. Conversely any term in $$S_2$$ with that property is a multiple of $$p_2$$ but not divisible by $$p_1$$, and hence has form $$p_2m$$ where $$p_1 \nmid m$$, therefore it is a term in $$S_2'$$. Thus $$S_2'$$ is the sub-series of $$S_2$$ defined by taking all $$n$$ with $$p_1 \nmid n$$ but $$p_2 \mid n$$. As in the previous step insert zeros into $$S_2'$$ to make it 'line up' with $$S_2$$ then subtract $$S_2'$$ term by term from $$S_2$$, leaving the original $$S_2$$ with all terms with $$p_2 \mid n$$ 'punched out' to zero. Now we can 'collapse' these punched out zero terms from the series to be left with : $$$$\left( 1 - \frac{1}{p_2^s} \right) \left( 1 - \frac{1}{p_1^s} \right) \zeta(s) = S_2 - S_2' = \sum_{p_1, p_2 \nmid n} \frac{1}{n^s} \label{eq:term-2} \tag{4}$$$$

ie. the original $$\zeta(s)$$ series with all $$n \in \mathbb{N}$$ with $$p_1 \mid n$$ or $$p_2 \mid n$$ 'sieved out'.

Equations (\ref{eq:term-1}) and (\ref{eq:term-2}) suggest there may be a general pattern for all $$k \in \mathbb{N}$$ : $$$$\frac{1}{a_k} \zeta(s) = \left( 1 - \frac{1}{p_k^s} \right) \cdots \left( 1 - \frac{1}{p_1^s} \right) \zeta(s) = S_k - S_k' = \sum_{p_1, \ldots, p_k \nmid n} \frac{1}{n^s} \label{eq:term-k} \tag{5}$$$$

ie. all $$n$$ divisible by any one of $$p_1, \ldots, p_k$$ have been 'sieved out' from $$\zeta(s)$$. (Note $$n = 1$$ always survives the sieve). We can use induction to prove (\ref{eq:term-k}) does indeed hold $$\forall k \in \mathbb{N}$$. Assuming (\ref{eq:term-k}) true for a given $$k$$ consider the case for $$k + 1$$ : $$\begin{eqnarray*} \left( 1 - \frac{1}{p_{k+1}^s} \right) \cdots \left( 1 - \frac{1}{p_1^s} \right) \zeta(s) & = & \left( 1 - \frac{1}{p_{k+1}^s} \right) (S_k - S_k') \\ & = & \sum_{p_1, \ldots, p_k \nmid n} \frac{1}{n^s} \; - \sum_{p_1, \ldots, p_k \nmid n} \frac{1}{(p_{k + 1} n)^s} \\ & = & S_{k + 1} - S_{k + 1}', \hspace{2em} \mbox{say} \end{eqnarray*}$$

Any term $$p_{k + 1}n$$ that appears in $$S_{k + 1}'$$ cannot be divisible by any of $$p_1, \ldots, p_k$$, and so $$S_{k + 1}'$$ is a sub-series of $$S_{k + 1}$$. Thus any term $$p_{k + 1}n$$ of $$S_{k + 1}'$$ appears in $$S_{k + 1}$$ and is not divisible by any of $$p_1, \ldots, p_k$$, but is divisible by $$p_{k + 1}$$. Conversely any term of $$S_{k + 1}$$ that has that property is a multiple of $$p_{k + 1}$$ but is not divisible by any of $$p_1, \ldots, p_k$$, and hence has form $$p_{k + 1}m$$ where $$m$$ is not divisible by any of $$p_1, \ldots, p_k$$ - hence this term appears in $$S_{k + 1}'$$. Thus $$S_{k + 1}'$$ is the sub-series of $$S_{k + 1}$$ defined by taking all the indices $$n$$ with $$p_1, \ldots, p_k \nmid n$$ but $$p_{k + 1} | n$$.

Now as before insert zeros into $$S_{k + 1}'$$ to make it 'line up' with $$S_{k + 1}$$ then subtract $$S_{k + 1}'$$ term by term from $$S_{k + 1}$$ - this leaves the original $$S_{k + 1}$$ with all terms having $$p_{k + 1} \mid n$$ 'punched out' to zero. Now 'collapse' the punched out zero terms from the series and we are left with :

$$\left( 1 - \frac{1}{p_{k+1}^s} \right) \cdots \left( 1 - \frac{1}{p_1^s} \right) \zeta(s) = S_{k + 1} - S_{k + 1}' = \sum_{p_1, \ldots, p_{k + 1} \nmid n} \frac{1}{n^s}$$

ie. we have now 'sieved out' from $$\zeta(s)$$ all $$n$$ divisible by any one of $$p_1, \ldots, p_{k + 1}$$, and this is the equation (\ref{eq:term-k}) re-stated for case $$k + 1$$, thus completing the induction.

Then in equation (\ref{eq:term-k}), noting term $$n = 1$$ appears on the right, and no term with $$2 \leq n \leq p_k$$ can appear on the right, for such $$n$$ must contain one of $$p_1, \ldots, p_k$$ in its prime decomposition, and using the absolute convergence of the series on the right (which follows from $$\zeta(s)$$ being absolutely convergent for $$\mathrm{Re}(s) > 1$$) we have :

$$\begin{eqnarray*} \left| \frac{1}{a_k} \zeta(s) - 1 \right| & \leq & \sum_{\substack{n \geq p_k + 1, \\ p_1, \ldots, p_k \nmid n}}^{\infty} \left| \frac{1}{n^s} \right| \\ & \leq & \sum_{n \geq p_k + 1}^{\infty} \left| \frac{1}{n^s} \right| \hspace{2em} \mbox{(since the former series is a sub-series of the latter)} \\ & \leq & \sum_{n \geq k + 1}^{\infty} \left| \frac{1}{n^s} \right| \\ & = & T - T_k \hspace{2em} \mbox{(the tail of the convergent series T = \displaystyle{\sum_{n = 1}^{\infty} \left| \frac{1}{n^s} \right|})} \\ & \rightarrow & 0 \mbox{ as } k \rightarrow \infty \\ \end{eqnarray*}$$

Hence etc, $$\hspace{30em}$$ QED