On the harmonic number ($H_n$) upper and lower "classical" bounds: which of those is closest to $H_n$? It is a well-known fact that the harmonic number
$$\displaystyle H_n = \sum_{k=1}^n \frac{1}{k}$$
satisfies the following inequality:
$$\displaystyle \ln(n) + \frac{1}{n} \;\leq \; H_n \; \leq \; \ln(n) + 1$$
as it is stated on page 26 of this notes.
Is it true that $H_n$ is closer to $\ln(n) + 1$ than $H_n$ is to $\displaystyle \ln(n) + \frac{1}{n}$? If so, how to prove that?
 A: 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
$$
H_n-\log(n)\le\gamma+\frac1{2n-1}\tag4
$$
A: 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.
A: 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
$$H_n>\log(n)+\frac12$$
