Introduction
I recently was playing around with some series and a natural question came to mind how many ways can a series equivalence be decomposed "a certain way" to a equivalence sequence?
Background
I believe for if the below holds for all integer $n$ holds then:
$$ \sum_{r=1}^n\frac{a_r}{r} = \frac{1}{n}\sum_{r=1}^n b_r \implies b_r = \sum_{r|l} a_l $$
Hence, in some sense we have managed to uniquely decompose the series. But if we sum the series $ \sum_{r=1}^n\frac{a_r}{r} = \frac{1}{n}\sum_{r=1}^n b_r $ on both sides then:
$$ \sum_{m}^n \sum_{r=1}^m\frac{a_r}{r} = \sum_{m}^n \frac{1}{m}\sum_{r=1}^m b_r $$
Simplifying the above expression:
$$ \implies \sum_{r=1} ^n(n-r+1) \frac{a_r}{r} = \sum_{r=1}^n b_r\Big((\sum_{k=1}^n \frac{1}{k}) - (\sum_{k=2}^r \frac{1}{k-1})\Big) $$
But there are now two ways (as far as I know) to decompose this series:
$$ \sum_{r=1}^n (n-r+1) \frac{a_r}{r} = \sum_{r=1}^n b_r\Big((\sum_{k=1}^n \frac{1}{k}) - (\sum_{k=2}^r \frac{1}{k-1})\Big) \implies b_r = \sum_{r|l} a_l $$
Or another decomposing solution is given by the manipulation in the equation with the L.H.S:
$$ (n+1)\sum_{r=1}^n \frac{a_r}{r} - \sum_{r=1}^n r \frac{a_r}{r} = \sum_{r=1}^n \frac{\tilde b_r}{n} + (n+1)\sum_{r=1}^n \frac{a_r}{r} = \sum_{r=1}^n b_r\Big((\sum_{k=1}^n \frac{1}{k}) - (\sum_{k=2}^r \frac{1}{k-1})\Big) $$
where $\tilde b_r = \sum_{r|l} l a_l $. Hence,
$$ \sum_{r=1}^n \frac{\tilde b_r}{n} + (n+1)\sum_{r=1}^n \frac{c_r}{n} = \sum_{r=1}^n b_r\Big((\sum_{k=1}^n \frac{1}{k}) - (\sum_{k=2}^r \frac{1}{k-1})\Big) $$
where $c_r =\sum_{r|l} a_l $
Setting each term to $0$ rather than their sum:
$$ \sum_{r|l} l a_l + (n+1) \sum_{r|l} a_l = n b_r\Big((\sum_{k=1}^n \frac{1}{k}) - (\sum_{k=2}^r \frac{1}{k-1})\Big) $$
Hence, this can also be decomposed as:
$$ \sum_{r|l} (1+l) a_l = \frac{n}{(n+1) } b_r\Big((\sum_{k=1}^n \frac{1}{k}) - (\sum_{k=2}^r \frac{1}{k-1})\Big) $$
Question
The number of ways of decomposing a series seems to be some function of $\deg(n)$ in the summation of the L.H.S? Can anyone prove what it is?