The harmonic sum is log x plus gamma plus O(1/x)

Since $t\mapsto 1\mathrm{/}t$ is decreasing on $[1,\mathrm{\infty })$ by The natural logarithm has derivative 1/x and equals the integral from 1 to x of 1/t, monotonicity of the integral and additivity over subintervals give ${\int }_n^{n+1}\frac{dt}{t}\le \frac{1}{n}\text{for every integer }n\ge 1,$ and $\frac{1}{n}\le {\int }_{n-1}^n\frac{dt}{t}\text{for every integer }n\ge 2.$

Put $E_N:={\sum }_{n=1}^{N}1\mathrm{/}n-\log (N+1)$. Summing the first inequality of step 1.1 from $n=1$ to $N$ gives $E_N\ge 0$, and the same inequality at $n=N+1$ gives $E_{N+1}-E_N=\frac{1}{N+1}-{\int }_{N+1}^{N+2}\frac{dt}{t}\ge 0.$ Also summing the second inequality of step 1.1 from $n=2$ to $N$ gives ${\sum }_{n=1}^N\frac{1}{n}\le 1+\log N<1+\log (N+1),$ so $0\le E_N<1$. Thus $(E_N)$ is increasing and bounded, hence convergent by A nondecreasing sequence bounded above converges to the supremum of its range, and a nonincreasing sequence bounded below to the infimum.

Let $L:={\lim }_N{E}_N$. Since ${\sum }_{n=1}^N\frac{1}{n}-\log N=E_N+\log \text{\hspace{0.17em}⁣}(1+\frac{1}{N}),$ and $0\le \log (1+1\mathrm{/}N)\le 1\mathrm{/}N$ by step 1.1, the sequence in The Euler-Mascheroni constant has the same limit $L$. Therefore $L=\gamma$, and $0\le {\sum }_{n=1}^N\frac{1}{n}-\log N-\gamma \le \frac{1}{N}\mathrm{.}$

For the given real $x$, let $N:=\lfloor x\rfloor$. Then $N\le x<N+1$, so $0\le \log x-\log N=\log (x\mathrm{/}N)\le \log (1+1\mathrm{/}N)\le 1\mathrm{/}N$. Combining this with step 3.1 yields ${\sum }_{n\le x}\frac{1}{n}={\sum }_{n=1}^N\frac{1}{n}=\log x+\gamma +O(1\mathrm{/}N)=\log x+\gamma +O(1\mathrm{/}x),$ because $N\ge x\mathrm{/}2$ for every $x\ge 1$.