# sequences of nonnegative reals such that a condition with series is satisfied

Find all sequences $$a_k$$ of nonnegative reals such that $$\sum a_k$$ converges and $$\sum_{k=1}^{\infty} a_{kn} = \frac{1}{\sqrt{n}}\sum_{k=1}^{\infty}a_k$$ for all $$n\in\mathbb{N}$$.

My friend asked me this question a month ago, I have tried but also without any result. Probably the sequence of 0's is the only one, but spending few hours without constructing a nontrivial example is not a proof the example does not exist.

There are no non-trivial solutions; the sequence of zeros is the only one.

How so? Well, let's see. Say, $$\sum_{k=1}^{\infty}a_k=S$$; then what is the sum of all terms in odd positions? Obviously,

$$\sum_{k=1}^{\infty}a_{2k-1}=S-\sum_{k=1}^{\infty}a_{2k}=\left(1-{1\over\sqrt2}\right)S$$

Now what is the sum of all $$a_k$$ where $$k$$ is neither divisible by $$2$$ nor by $$3$$? Well, we have to subtract the multiples of $$2$$, then of $$3$$, then add the multiples of 6 in the spirit of the inclusion–exclusion principle, otherwise those would be subtracted twice. What remains is

$$\left(1-{1\over\sqrt2}-{1\over\sqrt3}+{1\over\sqrt6}\right)S = \left(1-{1\over\sqrt2}\right)\cdot\left(1-{1\over\sqrt3}\right)S$$

Now let's throw out those divisible by 5. After more fiddling with inclusion–exclusion, we end up with

$$\left(1-{1\over\sqrt2}\right)\cdot\left(1-{1\over\sqrt3}\right)\cdot\left(1-{1\over\sqrt5}\right)S$$

By now we must have recognized the pattern, so let's just proceed to the end. On the LHS we have a lonely $$a_1$$, since any other $$k$$ than $$1$$ is divisible by some prime and hence gets kicked out at some step or another.

$$a_1=\prod_{n=1}^\infty\left(1-{1\over\sqrt p_n}\right)\cdot S$$

Now, a product $$\prod_{n=1}^\infty(1+x_n)$$ converges to a finite non-zero value iff the sum $$\sum_{n=1}^\infty x_n$$ converges, which in out case it obviously doesn't. Our product is $$0$$, and so is $$a_1$$.

Multiply all coefficients in this reasoning by $$2$$ to conclude the same about $$a_2$$, then proceed to the rest of the terms.

So it goes.