I've been obsessed with this one problem for many months now, and today is the sad day that I admit to myself I won't be able to solve it and I need your help. The problem is simple. We let

$$\mathbb{E}_{n\in\mathbb{N}}[f(n)]:=\lim_{N\to\infty}\frac{1}{N}\sum_{n=1}^{N}f(n)$$

denote the expected value of a function $f(n)$ (if it exists). This means that for some fixed choice of sequence $(a_n)_{n=1}^{\infty}$ the quantities $\mathbb{E}_{n\in\mathbb{N}}[a_{kn}]$ will give the "average value of every $k$" elements. For example, $\mathbb{E}_{n\in\mathbb{N}}[a_{3n}]$ will be the average of $a_{3},a_{6},a_{9}$ etc... . If all of these quantities exist, we will call a sequence $a_n$ to be "sequentially summable". The problem is as follows:

Show that for any bounded sequentially summable $a_n$ \begin{equation}\sum_{k=1}^{\infty}\frac{\lambda(k)\mathbb{E}_{n\in\mathbb{N}}[a_{kn}]}{k}=0\tag{1}\end{equation} where $\lambda(n)$ denotes the Liouville function

Swapping which side the term $\mathbb{E}_{n\in\mathbb{N}}[a_n]$ is on, this gives an absolutely lovely formula for $\mathbb{E}_{n\in\mathbb{N}}[a_n]$ from $\mathbb{E}_{n\in\mathbb{N}}[a_{kn}]$ ($k\geq 2$) which can be appreciated by even people who have done no mathematics. Similarly, I would expect that the conjecture holds for $\lambda(n)$ replaced by its "twin" $\mu(n)$, the Mobius function. Throughout the rest of this question, I will give all of the partial results and a general outline of how they are obtained.

This first partial results answers the question "why on earth would these values be 0???":

Partial Result $\#1$: For any bounded sequentially summable sequence $a_n$, it holds that \begin{equation}\lim_{s\to1^+}\sum_{k=1}^{\infty}\frac{\lambda(k)\mathbb{E}_{n\in\mathbb{N}}[a_{kn}]}{k^s}=\lim_{s\to1^+}\sum_{k=1}^{\infty}\frac{\mu(k)\mathbb{E}_{n\in\mathbb{N}}[a_{kn}]}{k^{s}}=0\tag{2}\end{equation} and thus if the sum in (1) converges it must converge to $0$

This result is obtained by inverting the summation order and exploiting the fact that $$\sum_{d|n}\frac{\mu(n)}{n^s}=\prod_{p}\left(1-\frac{1}{p^s}\right)>0$$ which means that the triangle inequality does not "lose" any sign cancelation. The following partial result is much stronger:

Partial Result $\#2$: There exists an absolute constant $c_0$ such that for any sequentially summable sequence $a_n$ and $N>0$ \begin{equation}\left|\sum_{k=1}^N\frac{\lambda(k)\mathbb{E}_{n\in\mathbb{N}}[a_{kn}]}{k}\right|<c_0m\tag{3}\end{equation} where $$m^2=\limsup_{N\to\infty}\frac{1}{N}\sum_{n=1}^Na^2_{n}$$

An outline of the proof of this result is given in this Math.SE question, but in essence the result comes from the fact that the boundedness of

\begin{equation}\left|\sum_{k=1}^N\frac{f(k)\mathbb{E}_{n\in\mathbb{N}}[a_{kn}]}{k}\right|\end{equation}

for some function $f(k)$ is essentially equivalent to tight enough bounds on the partial sums

$$\sum_{n=1}^{N}\frac{f(mk)}{k}$$

for all values of $m<N$, which $\mu(n)$ definitely has. The boundedness in Partial Result $\#2$ can be translated to bounds when $\mu(n)$ is replaced by $\lambda(n)$ by exploiting the identity $\lambda(n)=\sum_{d^2|n}\mu\left(\frac{n}{d^2}\right)$ and the relative uniformity of the bound w.r.t out choice of $a_n$. Here is the next partial result

Partial Result $\#3$: If, for any bounded sequentially summable sequence $a_n$, we have that \begin{equation} \lim_{N\to\infty}\frac{1}{N}\sum_{k=1}^{N}\lambda(k)\mathbb{E}_{n\in\mathbb{N}}[a_{kn}]=0 \end{equation} then our conjecute (1) holds.

This is a very important partial result since it is often much easier to show that the average of coefficients is zero than to show that they converge when summed with a factor of $\frac{1}{k}$. This result is done in a "smoothing" manner where Partial Result $\#3$ can be considered as control over short intervals and Partial Result $\#2$ can be considered as control over large intervals which we can bound as error.

At this point I will note that the condition that $a_n$ be bounded is quite important. For example, if $a_n=\Lambda(n)$ is the Von-Magnolt function then by the PNT $\mathbb{E}_{n\in\mathbb{N}}[a_n]=1$ but since $kn$ will have "few" prime powers for $k>1$ we have that $\mathbb{E}_{n\in\mathbb{N}}[a_{kn}]=0$ and thus our sum converges, but not to $0$.

I add as well that

$$\sum_{k=1}^{N}\frac{\mu(k)}{k}\frac{1}{[N/k]}\sum_{n=1}^{[N/k]}a_{kn}=\frac{a_1}{N}$$

due to simple inversion of summation order, and so since

$$\frac{1}{[N/k]}\sum_{n=1}^{[N/k]}a_{kn}\approx \mathbb{E}_{n\in\mathbb{N}}[a_{kn}]$$

we get further intuition for the result.