The boundedness of $1+\frac{1}{2}+\cdots+\frac{1}{n}-\log n$.

Question

As we all know, the sequence $1+\frac{1}{2}+\cdots+\frac{1}{n}-\log n$ is decreasing and bounded. Its limit is called Euler constant denoted by $\gamma$. I can get $$0<1+\frac{1}{2}+\cdots+\frac{1}{n}-\log n<1.$$ And if we use the limit $\gamma=0.577~215\cdots$, we know $$1+\frac{1}{2}+\cdots+\frac{1}{n}-\log n>\frac{1}{2}.$$ Now my question is how to prove $$1+\frac{1}{2}+\cdots+\frac{1}{n}-\log n>\frac{1}{2}$$ without the properties of limit and $\gamma=0.577~215\cdots$.

Answer 1

For integer $n > 1$, $\log n$ is the area below the curve $y = 1/x$ for $1 \le x \le n$. If we estimate this area by the trapezium rule, the result will be an over-estimate because the graph of $y = 1/x$ lies below its chords. Hence $$\log n < {1\over2}\left({1\over1}\right) + {1\over2} + {1\over 3} + \cdots + {1\over{n-1}} + {1\over2}\left({1\over n}\right),$$ which gives $$1 + {1\over 2} + \cdots + {1\over n} - \log n > {1\over 2} + {1\over 2n}.$$

Answer 2

Any other elementary proof will welcome.

I get an idea from Showing $\gamma < \sqrt{1/3}$ without a computer

$$\gamma = \int_{0}^{+\infty}\left(\frac{1}{1-e^{-x}}-\frac{1}{x}\right)e^{-x}\,dx.\tag{1}$$ Since $f(x)=\frac{1}{1-e^{-x}}-\frac{1}{x}$ is concave on $\mathbb{R}^+$, for any $x\in\mathbb{R}^+$ we have $f(x)>\frac{1}{2}$, so: $$\gamma>\int_{0}^{+\infty}\frac{1}{2}e^{-x}\,dx = \frac{1}{2}.$$ By the same way, $f(x)>\frac{1}{2}+\frac{1}{2}x-\frac{1}{720}x^3$, $$\gamma>\int_{0}^{+\infty}\left(\frac{1}{2}+\frac{1}{2}x-\frac{1}{720}x^3\right)e^{-x}\,dx = \frac{23}{40}=0.575.$$ The Taylor expansion of $f(x)$.