I see on that this post that the inequality $\log{n} < H_n < \log{n} + 1$ is proven. However, Proofs from THE BOOK takes it a step further, and states that $\log{n} + \frac{1}{n} < H_n < \log{n} + 1$. Its explanation is as follows:
Let
$$H_n = \sum_{k=1}^n \frac{1}{k}.$$
Consider the following image:
From it, we can derive that,
$$H_n - 1 = \sum_{k=2}^n \frac{1}{k} < \int_1^n \frac{1}{t}dt = \log{n}$$
by comparing the area below the graph of $f(t) = \frac{1}{t} (1 \leq t \leq n)$ with the area of the dark shaded rectangles, and
$$H_n - \frac{1}{n} = \sum_{k=1}^{n-1} \frac{1}{k} > \int_1^n \frac{1}{t}dt = \log{n}$$
by comparing the area of the large (including the lightly shaded parts). Taken together, this yields,
$$\log{n} + \frac{1}{n} < H_n < \log{n} + 1.$$
I understand where the $\log{n} + 1$ comes from (it is explained in the aforementioned post on the same topic), but I don't understand the $\log{n} + \frac{1}{n}$, nor its explanation. Can anyone help?
The image depicts $n-1$ large rectangles, each with width $1$. The first rectangle is placed over the interval $[1,2]$, the second over the interval $[2,3]$, the final one over the interval $[n-1,n]$. Their corresponding heights are $\frac11$, $\frac12,\ldots,\frac1{n-1}$. Hence their total area is $$\frac11+\frac12+\cdots+\frac1{n-1}=\sum_{k=1}^{n-1}\frac1k.$$ By inspection, the total area of these large rectangles exceeds the area under the curve $y=\frac1t$ from $t=1$ to $t=n$: $$\sum_{k=1}^{n-1}\frac1k > \int_{t=1}^{t=n}\frac1t\,dt.$$ The assertion $H_n-\frac1n = \sum_{k=1}^{n-1}\frac1k$ follows from the definition of $H_n$; the assertion $\int_1^n\frac1t\,dt=\log n$ is calculus. Put everything together to conclude $H_n -\frac1n > \log n$.