Question
So I conjectured a formula which was proven:
Let $b_r = \sum_{d \mid r} a_d\mu(\frac{m}{d})$. We prove that if the $b_r$'s are small enough, the result is true.
Claim: If $\lim_{n \to \infty} \frac{\log^2(n)}{n}\sum_{r=1}^n |b_r| = 0$ and $f$ is smooth, then $$\lim_{k \to \infty} \lim_{n \to \infty} \sum_{r=1}^n a_rf\left(\frac{kr}{n}\right)\frac{k}{n} = \left(\lim_{s \to 1} \frac{1}{\zeta(s)}\sum_{r=1}^\infty \frac{a_r}{r^s}\right)\int_0^\infty f(x)dx.$$
My question is what is the cardinality of the set of $a_r$?
Reason for confusion
Focusing on the L.H.S
This seems to say for every point on the curve can be mapped to $f(x)$ which in turn can be mapped to a coefficient $a_r$ .
$$ x \to f(x) \to a_r $$
Hence, the set has cardinality $ 2^{\aleph_0} $
Focusing on the R.H.S
This seems to say the number of $a_r$ must must be the same as that of the natural numbers.
Hence, the set has cardinality $ \aleph_0 $
The indices $r$ of the sequence $\{a_r\}$ are natural numbers. Thus $\{a_r\}$ is a sequence of non-negative integers (or, more general, real) numbers. This sequence is predefined by the condition that $a_r$ is the weight (the number of recounts) of $r$-th strip $S_{n,k}$ in the sum $\sum_{r=1}^n a_rf\left(\tfrac{k}{n}r\right)\tfrac{k}{n}$. But strip $S_{n,k}$ depends on $n$ and $k$, and a value $f\left(\tfrac kn r\right)$, corresponding to $S_{n,k}$, depends on $n$ and $k$ too. So to a coefficient $a_r$ corresponds not one strip, and even not a sequence of strips tending to a segment from $(x,0)$ to $(x,f(x))$ for some $x=x(r)$. To $a_r$ corresponds a two-parametric family $\{S_{n,k}\}$ of stips, where $S_{n,k}$ is a strip from $(\tfrac{k}{n}r,0)$ to $(\tfrac{k}{n}r, f\left(\tfrac kn r\right))$ of width $\tfrac{k}{n}$. Thus I don’t see a natural (and, moreover, bijective) map, mapping points $(x,f(x))$ on the graph of $f$ to some $r$.