2
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.