Proving that $\sum_{n = 1}^{\infty} \frac{1}{n^s} = \prod_{k=1}^{\infty}\frac{1}{1-p_k^{-s}}$

Could you please have a look at my solution of the following exercise?

Show that for $$s>1$$ holds $$\sum_{n = 1}^{\infty} \frac{1}{n^s} = \prod_{k=1}^{\infty}\frac{1}{1-p_k^{-s}}$$

where $$(p_k)_k$$ is the increasing sequence of primes.

I came up with the following solution:

At first we note that

$$\prod_{k=1}^{\infty}\frac{1}{1-p_k^{-s}} = \prod_{k=1}^{\infty} \frac{p_k^s}{p_k^s-1} = \prod_{k=1}^{\infty} \biggl( \sum_{j=0}^{\infty} \frac{1}{p_k^{s \cdot j}}\biggr)$$

Now we show both $$\ge$$ and $$\le$$:

$$\ge:$$

$$\prod_{k=1}^{N}\frac{1}{1-p_k^{-s}} = \prod_{k=1}^{N} \frac{p_k^s}{p_k^s-1}$$

To make things easier we take care that the exponents are natural numbers

$$\le \prod_{k=1}^{N} \frac{p_k^{\lceil s \rceil}}{p_k^{\lfloor s \rfloor} -1}$$

For fitting natural numbers $$A,B$$ we may write

$$= \frac{A}{B} = \sum_{n=1}^{A} \frac{1}{B}$$

And since $$A \ge B$$

$$\le \sum_{n=1}^{A} \frac{1}{n}$$

$$\le:$$

We make the observation that in the result of the multiplication

$$\biggl( \sum_{i=0}^{a} \frac{1}{p_1^i} \biggr) \cdot \biggl( \sum_{j=0}^{b} \frac{1}{p_2^j}\biggr) = \sum_{i=0}^{a} \sum_{j=0}^{b} \frac{1}{p_1^i \cdot p_2^j}$$

each combination of powers of $$p_1$$ and $$p_2$$ up to $$p_1^a$$ and $$p_2^b$$ appears exactly once in the denominators on the right side above. By generalising this and remembering about the fundametal theorem of arithmetic we obtain:

$$\sum_{n = 1}^{N} \frac{1}{n^s} \le \prod_{k=1}^{A} \biggl( \sum_{j=0}^{B_k} \frac{1}{p_k^{s \cdot j}}\biggr)$$

for some natural numbers $$A$$ and $$B$$ and further:

$$\le \prod_{k=1}^{A} \biggl( \sum_{j=0}^{\infty} \frac{1}{p_k^{s \cdot j}}\biggr) = \prod_{k=1}^{A} \frac{p_k^s}{p_k^s-1} \le \prod_{k=1}^{\infty} \frac{p_k^s}{p_k^s-1}$$

Is this correct and if yes, could you tell me where exactly we need that $$s>1$$; I do not see why this is required.

• Let $A_K$ be the set of integers whose prime factors are all $\le p_K$. Then for $s > 0$, $\prod_{k=1}^K \frac{1}{1-p_k^{-s}} = \prod_{k=1}^K (1+ \sum_{j=1}^\infty (p_k^j)^{-s}) = \sum_{n \in A_K} n^{-s}$. For $s > 1$ then $\sum_{n=1}^\infty n^{-s} - \prod_{k=1}^K \frac{1}{1-p_k^{-s}}=\sum_{n \not \in A_K}n^{-s}\le \sum_{n=P_K+1}^\infty n^{-s}$ whence $\lim_{K \to \infty}\sum_{n=1}^\infty n^{-s}-\prod_{k=1}^K\frac{1}{1-p_k^{-s}} = 0$. For $s\in (0,1]$ then $\lim_{K \to \infty}\prod_{k=1}^K\frac{1}{1-p_k^{-s}}=\lim_{K \to \infty}\sum_{n\in A_K}n^{-s}=\sum_{n=1}^\infty n^{-s}= \infty$ – reuns Jan 6 at 17:36
• "could you tell me where exactly we need that $s>1$" Actually, we do not. When $s\leqslant1$, the identity in the question holds as well, but it reads $\infty=\infty$. – Did Jan 6 at 17:54
• – Robert Z Feb 14 at 11:45