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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:

this

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?

Baguette Boy
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  • if we have $g(x) > 0$ but $g'(x) < 0,$ then $$ \int_a^{b+1} ; g(x) ; dx ; < ; \sum_{j=a}^b ; g(j) ; < ; \int_{a-1}^b ; g(x) ; dx $$ – Will Jagy Jan 05 '23 at 22:18
  • the $\log(n)+\frac{1}{n}$ comes from the second inequality after moving the $\frac{1}{n}$ to the other side. The inequality comes from the upper Riemann sum and since $$\int=\sup{\Sigma_U}$$, where $ \Sigma_U$ is a upper sum, the inequality follows trivially – Fix Jan 05 '23 at 22:20
  • $$ \int_1^{n+1} ; \frac{1}{x} ; dx ; < ; \sum_{j=1}^n ; \frac{1}{j} ; < ; \int_{0}^n ; \frac{1}{x} ; dx $$ but also $$ \int_2^{n+1} ; \frac{1}{x} ; dx ; < ; \sum_{j=2}^n ; \frac{1}{j} ; < ; \int_{1}^n ; \frac{1}{x} ; dx $$ and $$1 + \int_2^{n+1} ; \frac{1}{x} ; dx ; < ; \sum_{j=1}^n ; \frac{1}{j} ; < ; 1+ \int_{1}^n ; \frac{1}{x} ; dx $$ so $$ 1 - \log 2 + \log(n+1) < H_n < 1 + \log n $$ – Will Jagy Jan 05 '23 at 22:22
  • Notice that the sum extends to $n-1$, so we are missing the term $1/n$ – David Raveh Jan 05 '23 at 22:22
  • $$ 1 - \log 2 + \log(n+1) < H_n < 1 + \log n $$ and $$ 1 - \log 2 + \log(1 + \frac{1}{n}) < H_n - \log n < 1 $$ – Will Jagy Jan 05 '23 at 22:33

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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 $H_n -\frac1n > \log n$.

grand_chat
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