Solution September 20, 2007


Does there exist a sequence \{b_{i}\}_{i=1}^\infty of positive real numbers such that for each natural m: b_{m}+b_{2m}+b_{3m}+\dots=\frac1m


Assume that there is such a sequence (bi).

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


S_P := \sum_{P \nmid i}b_i

SP is well-defined because it sums a subsequence of (bi) and all bi are positive.

Lemma: If P is finite, then

S_P = \prod_{p \in P} \left(1-\frac1p\right)

Proof: By induction on the size of P. If P=\empty then S_P = \sum_{i=1}^\infty b_i = 1.

Now let P = Q \cup\{p\} with p\not\in Q. Consider the sequence c_i := p\cdot b_{pi}. It satisfies all requirements of (bi). Therefore,

\sum_{Q\nmid n}c_i = \sum_{Q\nmid n} b_i = \prod_{p \in Q} \left(1-\frac1p\right).

A number n satisfies Q\nmid n if either P\nmid n or p | n and Q\nmid \frac np. So

\sum_{P\nmid i} b_i = \sum_{Q\nmid i} b_i - \frac1p\sum_{Q\nmid i} c_i

and we get

\sum_{P\nmid i} b_i = \prod_{p \in P} \left(1-\frac1p\right)

This completes the induction step and thus proves the lemma.

Let P_k = \{p | p \leq k, p \mathrm{\ prime}\}.

Because S_{P_k} \leq b_1 + \sum_{i=k}^\infty b_i we have that

\lim_{k\to\infty}S_{P_k} = b_1


b_1 = \prod_{p \mathrm{\ prime}}\left(1-\frac1p\right) = 0

which is not positive. Thus no sequence (bi) satisfying the problem's requirements exists.

Note: By considering c_i := k\cdot b_{ki} we can show that bk = 0 for all k. So the answer does not change if we replace "positive" by "nonnegative".

Short "proof"

This "proof" makes use of the Möbius function (

Assume there is such a sequence (bi). Let c_m = b_{m}+b_{2m}+b_{3m}+\dots

Consider the sum

S = \sum_{m=1}^\infty \mu(m)c_{m}

Now on the one hand,

S = \sum_{m=1}^\infty \frac{\mu(m)}m = 0

On the other hand,

S = \sum_{m=1}^\infty\sum_{i=1}^\infty \mu(m)b_{im} = \sum_{n=1}^\infty\sum_{d|n}\mu(d)b_{n} = b_1\,

So b1 = 0 which doesn't satisfy the requirements of the problem.