# Sum of reciprocals of squarefree numbers

• Jun 15th 2009, 09:55 PM
Bruno J.
Sum of reciprocals of squarefree numbers
This one is for fun. I'll give my solution later.
Let $\displaystyle S$ be the set of squarefree positive integers. Show, as simply as you, can that $\displaystyle \sum_{s \in S} \frac{1}{s} = \infty$.

Of course an instant solution is given by the fact that the sum of the reciprocals of the primes diverges. However try using another way for fun!
• Jun 19th 2009, 12:04 PM
Bruno J.
Anyways here is my solution.
(1) every integer can be uniquely represented by the product of a square and a square-free number.
(2) $\displaystyle \sum_{j=1}^\infty\frac{1}{j^2}$ converges. This is easy to show and we don't need to know what it converges to.

But by (1) we have that

$\displaystyle \Big(\sum_{j=1}^\infty\frac{1}{j^2}\Big)\Big(\sum_ {s\in S} \frac{1}{s}\Big)$

is the harmonic series; hence $\displaystyle \sum_{s\in S} \frac{1}{s}$ diverges.
• Jun 20th 2009, 10:19 AM
Bruno J.
This can be generalized as follows. Suppose we have $\displaystyle n$ sets of integers, $\displaystyle S_1,...,S_n$, such that each integer can be uniquely represented as a product $\displaystyle s_1...s_n$. Then one of $\displaystyle \sum_{s \in S_j}\frac{1}{s}$ diverges.

This is quite interesting because if we have an infinite family of sets $\displaystyle S_0,S_1,...$ instead of a finite one then the above does not necessarily hold anymore! For instance if we take the nth set to be the set of powers of the nth prime (so that we have the usual representation as a product of prime powers), then all of the sums converge (they're just geometric series).