Asymptotic of $\sum_{n = 1}^{N}\frac{1}{n}$ versus $\sum_{1 \leq n \leq x}\frac{1}{n}$ Consider the series $\sum_{n = 1}^{N}\frac{1}{n}$. It is well known that we have the asymptotic: $$\sum_{n = 1}^{N}\frac{1}{n} = \log N + \gamma + \frac{1}{2N} + \frac{1}{12N^{2}} + O(N^{-3}).$$
My question: Consider the similar sum $F(x) = \sum_{1 \leq n \leq x}\frac{1}{n}$. Note that $F(N) = \sum_{n = 1}^{N}\frac{1}{n}$. Then does $F$ has the same asymptotic above? That is, is
$$F\sum_{1 \leq n \leq x}\frac{1}{n} = \log x + \gamma + \frac{1}{2x} + \frac{1}{12x^{2}} + O(x^{-3})?$$
I think this should be true, but the only thing I can come up with is to look at $F(\lfloor x \rfloor)$ and use the asymptotic for $\sum_{n = 1}^{\lfloor x \rfloor}\frac{1}{n}$ but I am unsure how to compare $\log x$ and $\log \lfloor x \rfloor$ in a way that the error that occurs is $O(x^{-3})$.

Note that $$\log x - \log \lfloor x \rfloor=\log x - \log\left(x - [x]\right)=-\log\left(1-\frac{[x]}{x}\right)=\frac{[x]}{x}+O(x^{-2}),$$
where $[x]$ is the fractional part of $x$, and is clearly $O(1)$.  That is, the oscillations around the smooth curve $\log x$ contribute already at the same order as $1/x$.  Therefore, the best you can say is that
$$
F(x)=\log x + \gamma + O(x^{-1}).
$$
