I am studying asymptotic expansion of the sum $\displaystyle\sum_{n \le x}\cfrac{1}{n}$ and I need some help to clarify an argument as below:
The claim is that $\displaystyle\sum_{n \le x}\cfrac{1}{n} = \log x + C + O(\frac{1}{x})$ where $C = \displaystyle \lim_{x\to \infty}(\sum_{n \le x}\cfrac{1}{n} -\log x) $ is the Euler's constant.
Using Euler's summation formula, taking $f(t) =\cfrac{1}{t}$ we have
$\displaystyle\sum_{n \le x}\cfrac{1}{n} = \int_1^x\cfrac{dt}{t}-\int_1^x\cfrac{t-[t]}{t^2}dt+1-\cfrac{x-[x]}{x}=\int_1^x\cfrac{dt}{t}-\int_1^x\cfrac{t-[t]}{t^2}dt+1 + O(\cfrac{1}{x})$.
Then, the book does the following trick for $\displaystyle\int_1^x\cfrac{t-[t]}{t^2}dt$:
$\displaystyle\int_1^x\cfrac{dt}{t}-\int_1^x\cfrac{t-[t]}{t^2}dt+1 + O(\cfrac{1}{x}) = \log x + 1 -\int_1^\infty\cfrac{t-[t]}{t^2}dt + \int_x^\infty\cfrac{t-[t]}{t^2}dt. $
Then says $\displaystyle\int_1^\infty\cfrac{t-[t]}{t^2}dt$ is dominated by $\displaystyle\int_1^\infty\cfrac{1}{t^2}dt$ and $0 \le \displaystyle\int_x^\infty\cfrac{t-[t]}{t^2} dt \le \displaystyle\int_x^\infty\cfrac{1}{t^2} = \cfrac{1}{x}dt$.
Finally,
$\displaystyle\sum_{n \le x}\cfrac{1}{n} = \log x + 1 -\int_1^\infty\cfrac{t-[t]}{t^2}dt + O(\cfrac{1}{x})$
and somehow saying that $\displaystyle 1 -\int_1^\infty\cfrac{t-[t]}{t^2}dt = C = \lim_{x\to \infty}(\sum_{n \le x}\cfrac{1}{n} -\log x) $ it concludes that
$\displaystyle\sum_{n \le x}\cfrac{1}{n} = \log x + C + O(\frac{1}{x})$ .
So, I want to ask:
1) Why do we need the "trick" above?
2) How $\displaystyle 1 -\int_1^\infty\cfrac{t-[t]}{t^2}dt$ can give the Euler's constant?
Thank you.
1) You do not need such trick to prove the statement. For any $n\geq 1$ we have that $\frac{1}{n}-\log\left(1+\frac{1}{n}\right)$ is bounded beween $\frac{1}{4n^2}$ and $\frac{1}{2n^2}$, hence $$\begin{eqnarray*} H_N = \sum_{n=1}^{N}\frac{1}{n} &=& \sum_{n=1}^{N}\log\left(1+\frac{1}{n}\right)+\sum_{n=1}^{N}\left[\frac{1}{n}-\log\left(1+\frac{1}{n}\right)\right]\\&=&\log(N+1)+\sum_{n\geq 1}\left[\frac{1}{n}-\log\left(1+\frac{1}{n}\right)\right]-O\left(\frac{1}{N}\right)\\&=&\log(N)+\gamma+O\left(\frac{1}{N}\right)\end{eqnarray*} $$ where $\gamma=\sum_{n\geq 1}\left[\frac{1}{n}-\log\left(1+\frac{1}{n}\right)\right]$ is the classical series representation for the Euler-Mascheroni constant. Since $\frac{1}{n}=\int_{0}^{+\infty}e^{-nx}\,dx$ and by Frullani's theorem $\int_{0}^{+\infty}\frac{e^{-nx}-e^{-(n+1)x}}{x}\,dx=\log\left(1+\frac{1}{n}\right)$, such series representation also gives an integral representation: $$ \gamma = \int_{0}^{+\infty}\left(\frac{1}{e^x-1}-\frac{1}{x e^x}\right)\,dx =\int_{0}^{1}\left(\frac{1}{\log(1-x)}+\frac{1}{x}\right)\,dx.$$ This is practical for estimating the magnitude of $\gamma$ through Gregory coefficients.
2) Because your manipulation proves that $$ \lim_{N\to +\infty}\left(H_N-\log N\right) = 1-\int_{1}^{+\infty}\frac{\{t\}}{t^2}\,dt,$$ hence $\gamma=1-\int_{1}^{+\infty}\frac{\{t\}}{t^2}\,dt$, the second term of the asymptotic expansion of $H_N$.