I have the sum ( $M$ is any integer $> 1$ ):
$$ \sum_{h = 1}^{M}\left(\,\left\lfloor\, 2M + 1 \over h\,\right\rfloor -\left\lfloor\, 2M \over h\,\right\rfloor\,\right) $$
and looking for a way to simplify it. In the sense of either finding a simple closed form or a good approximation for M large. This resembles my previous question involving the divisor summatory function. However, now this is different because the sum extends to $M$ (not $2M$) and now we have differences with at the numerators an odd number and the previous even number (which is $2$ times $M$), I was hoping some good simplification could be found in this case. The first terms are integers, so they pose no problems, I was mainly looking for some way to simplify the other differences.
A tight upper bound would also be useful ( as well as references to similar well-known $\mbox{formulas )}$.
I think you might have a typo based on your explanation. As written, this difference is nonzero if and only if $h|2M+1$. In that case the difference is $1$. Thus This is simply the number of divisors of $2M+1$, which is a well studied problem. Specifically, if $f(M)$ is the sum that you are requesting, then $f(M)=\sigma_0(2M+1)-1$.