Inequality involving sum with parameter in denominator (leading to harmonic number problem)

Given integers $a$, $b$ and $c$, I am trying to find $n\le b-1$ such that this inequality holds :

$\sum _{i=0}^{n}\frac{a}{b-i}\le c\le \sum _{i=0}^{n+1}\frac{a}{b-i}$

That is to say : the largest $n$ up to which the sum is the closest to $c$

Since :

$\sum _{i=0}^{n}\frac{a}{b-i}=a\cdot (\sum _{i=0}^{b}\frac{1}{i}-\sum _{i=0}^{b-n-1}\frac{1}{i})$

And $\sum _{i=0}^{k}\frac{1}{i}={H}_{k}$ where ${H}_{k}$ is the $k$-th harmonic number,

We can rewrite the inequality into :

$a({H}_{b}-{H}_{b-n-1})\le c\le a({H}_{b}-{H}_{b-n-2})$

How to find a tight lower and upper bound for $n$, such that this inequality holds ?

Given integers $a$, $b$ and $c$, I am trying to find $n\le b-1$ such that this inequality holds :

$\sum _{i=0}^{n}\frac{a}{b-i}\le c\le \sum _{i=0}^{n+1}\frac{a}{b-i}$

That is to say : the largest $n$ up to which the sum is the closest to $c$

Since :

$\sum _{i=0}^{n}\frac{a}{b-i}=a\cdot (\sum _{i=0}^{b}\frac{1}{i}-\sum _{i=0}^{b-n-1}\frac{1}{i})$

And $\sum _{i=0}^{k}\frac{1}{i}={H}_{k}$ where ${H}_{k}$ is the $k$-th harmonic number,

We can rewrite the inequality into :

$a({H}_{b}-{H}_{b-n-1})\le c\le a({H}_{b}-{H}_{b-n-2})$

How to find a tight lower and upper bound for $n$, such that this inequality holds ?