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.

  • 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.