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.
 A: 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.
