Vrftsjtryk

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:

$$ sum_{r=1}^nfrac{a_r}{r} = frac{1}{n}sum_{r=1}^n b_r implies b_r = sum_{r|l} a_l $$

Hence, in some sense we have managed to uniquely decompose the series. But if we sum the series $ sum_{r=1}^nfrac{a_r}{r} = frac{1}{n}sum_{r=1}^n b_r $ on both sides then:

$$ sum_{m}^n sum_{r=1}^mfrac{a_r}{r} = sum_{m}^n frac{1}{m}sum_{r=1}^m b_r $$

Simplifying the above expression:

$$ implies sum_{r=1} ^n(n-r+1) frac{a_r}{r} = sum_{r=1}^n b_rBig((sum_{k=1}^n frac{1}{k}) - (sum_{k=2}^r frac{1}{k-1})Big) $$

But there are now two ways (as far as I know) to decompose this series:

$$ sum_{r=1}^n (n-r+1) frac{a_r}{r} = sum_{r=1}^n b_rBig((sum_{k=1}^n frac{1}{k}) - (sum_{k=2}^r frac{1}{k-1})Big) implies b_r = sum_{r|l} a_l $$

Or another decomposing solution is given by the manipulation in the equation with the L.H.S:

$$ (n+1)sum_{r=1}^n frac{a_r}{r} - sum_{r=1}^n r frac{a_r}{r} = sum_{r=1}^n frac{tilde b_r}{n} + (n+1)sum_{r=1}^n frac{a_r}{r} = sum_{r=1}^n b_rBig((sum_{k=1}^n frac{1}{k}) - (sum_{k=2}^r frac{1}{k-1})Big) $$

where $tilde b_r = sum_{r|l} l a_l $. Hence,

$$ sum_{r=1}^n frac{tilde b_r}{n} + (n+1)sum_{r=1}^n frac{c_r}{n} = sum_{r=1}^n b_rBig((sum_{k=1}^n frac{1}{k}) - (sum_{k=2}^r frac{1}{k-1})Big) $$

where $c_r =sum_{r|l} a_l $

Setting each term to $0$ rather than their sum:

$$ sum_{r|l} l a_l + (n+1) sum_{r|l} a_l = n b_rBig((sum_{k=1}^n frac{1}{k}) - (sum_{k=2}^r frac{1}{k-1})Big) $$

Hence, this can also be decomposed as:

$$ sum_{r|l} (1+l) a_l = frac{n}{(n+1) } b_rBig((sum_{k=1}^n frac{1}{k}) - (sum_{k=2}^r frac{1}{k-1})Big) $$

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?