On the harmonic number ($H_n$) upper and lower "classical" bounds: which of those is closest to $H_n$?

Note that since $\frac1x$ is convex, then we have

$$\log(n)=\int_1^n\frac1x\,dx<\frac12+\sum_{k=2}^{n-1}\frac1k+\frac1{2n}\tag1$$

Rearranging $(1)$ reveals

$$H_n-\log(n)>\frac12+\frac1{2n}>\frac12$$

Therefore, we have

$$H_n>\log(n)+\frac12$$

Consider the sequence $H_n-\log(n)$. Jensen's Inequality says $$ \begin{align} (H_n-\log(n))-(H_{n+1}-\log(n+1)) &=\log\left(1+\frac1n\right)-\frac1{n+1}\\ &=\int_n^{n+1}\frac1x\,\mathrm{d}x-\frac1{n+1}\\ &\ge\left(\int_n^{n+1}x\,\mathrm{d}x\right)^{-1}-\frac1{n+1}\\ &=\frac1{n+\frac12}-\frac1{n+1}\\ &=\frac1{(2n+1)(n+1)}\\ &\ge\frac1{(2n+1)\left(n+\frac32\right)}\\ &=\frac1{2n+1}-\frac1{2n+3}\tag1 \end{align} $$ Thus, $H_n-\log(n)$ is a decreasing sequence and the limit is $\gamma$, the Euler-Mascheroni Constant.

Furthermore, summing $(1)$ yields $$ H_n-\log(n)\ge\gamma+\frac1{2n+1}\tag2 $$ As shown in this answer, $\gamma=0.57721566490153286060651209$. Thus, $H_n-\log(n)$ is closer to $1$ than to $\frac1n$ for $n\ge2$.

---

We can get an upper bound on $H_n-\log(n)$ as follows $$ \begin{align} (H_n-\log(n))-(H_{n+1}-\log(n+1)) &=\log\left(1+\frac1n\right)-\frac1{n+1}\\ &=\int_0^{1/n}\frac{x}{(1+x)^2}\,\mathrm{d}x\\ &\le\int_0^{1/n}x\,\mathrm{d}x\\ &=\frac1{2n^2}\\ &\le\frac1{2n-1}-\frac1{2n+1}\tag3 \end{align} $$ Summing $(3)$ yields $$ H_n-\log(n)\le\gamma+\frac1{2n-1}\tag4 $$

I finally got a proof. The area under $1/x$ from 1 to $n$

$\;\displaystyle \int_{1}^n \frac{1}{x}\, dx = \ln n\;$

could be bounded by using the trapezoidal rule. As a result,

$\displaystyle \sum_{k=1}^{n-1} \frac{1}{2}\left(\frac{1}{k} + \frac{1}{k+1}\right)1 \; \; > \;\; \ln n$

which implies that

$\displaystyle H_n \;\; > \;\; \ln n \,+\, \frac{1}{2} \,+\, \frac{1}{2n}$

by using the definition of $H_n$ in the summation.