# Solution September 20, 2007

### Problem

Does there exist a sequence of positive real numbers such that for each natural m:

### Solution

Assume that there is such a sequence (*b*_{i}).

Let *P* be a set of prime numbers. Say that *P* divides
*n* (symbolically, *P* | *n*) if *any* member of *P* divides *n*.

Define

*S*_{P} is well-defined because it sums a subsequence of (*b*_{i}) and all *b*_{i} are positive.

**Lemma:** If *P* is finite, then

**Proof:** By induction on the size of *P*. If
then .

Now let with . Consider the
sequence . It satisfies all requirements of (*b*_{i}). Therefore,

- .

A number *n* satisfies if either or *p* | *n* and . So

and we get

This completes the induction step and thus proves the lemma.

Let .

Because we have that

Therefore,

which is not positive. Thus no sequence (*b*_{i}) satisfying the problem's requirements exists.

**Note:** By considering we can show that *b*_{k} = 0 for all *k*. So the answer does not change if we replace "positive" by "nonnegative".

### Short "proof"

This "proof" makes use of the MÃ¶bius function (*http://mathworld.wolfram.com/MoebiusFunction.html*).

Assume there is such a sequence (*b*_{i}). Let

Consider the sum

Now on the one hand,

On the other hand,

So *b*_{1} = 0 which doesn't satisfy the requirements of the problem.