3
$\begingroup$

$$\sum_{k=0}^{\infty}\frac{1}{2^{2k}}\sum_{k=0}^{\infty}\frac{1}{3^{2k}}\sum_{k=0}^{\infty}\frac{1}{5^{2k}}\cdots=\sum_{n=1}^{\infty}\frac{1}{n^2}$$

Ignoring problems connected with convergence and rearrangement of terms, and the denominators on the left are the even powers of the primes.

  • 2
    HINT: [Fundamental theorem of arithmetic](http://en.wikipedia.org/wiki/Fundamental_theorem_of_arithmetic).2011-10-24
  • 0
    Could you make it clearer...2011-10-24
  • 0
    Keep in mind the geometric series when viewing http://en.wikipedia.org/wiki/Proof_of_the_Euler_product_formula_for_the_Riemann_zeta_function2011-10-24
  • 0
    Could someone give me a complete proof? I am still not very clear...2011-10-24

1 Answers 1

5

Suppose you have an arbitrary finite set of primes, e.g. $\{2,3,5\}$. Look at $$ \left(1 + \frac{1}{2^2} + \frac{1}{2^4} + \frac{1}{2^6} +\cdots \right)\left(1 + \frac{1}{3^2} + \frac{1}{3^4} + \frac{1}{3^6} +\cdots \right)\left(1 + \frac{1}{5^2} + \frac{1}{5^4} + \frac{1}{5^6} +\cdots \right). $$ This expands to $$ \sum_{a,b,c \ge 0} \frac{1}{2^{2a}} \frac{1}{3^{2b}} \frac{1}{5^{2c}} = \sum_{a,b,c \ge 0} \left(\frac{1}{2^a} \frac{1}{3^b} \frac{1}{5^c}\right)^2. $$ The fraction runs throught the list of reciprocals of all possible products of $2$s, $3$s, and $5$s.

Now instead of just those primes, $2$, $3$, and $5$, use the list of all primes. The set of all possible products of not-necessarily distinct primes is just the set of all positive integers.