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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?

PNT
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Diego S. Rodrigues
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3 Answers3

15

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

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

Mark Viola
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  • Thank you, Mark Viola, but I didn't understand your first inequality. Why doest the fact that $\displaystyle \frac{1}{x}$ is convex implies it? – Diego S. Rodrigues Nov 23 '17 at 17:54
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    @DiegoS.Rodrigues: by the Hermite-Hadamard inequality,if we approximate the integral of a positive and convex function through the trapezoid method we get an upper bound on the actual value of the integral. – Jack D'Aurizio Nov 23 '17 at 18:05
  • Yes, this helped me, but the Hermite-Hadamard Inequality gives $\displaystyle \int_{1}^{n} \frac{1}{x} dx ; \leq ; (n-1) \displaystyle \left(\frac{1}{2}+\frac{1}{2n}\right)$ as a result, and I couldn't proceed to get the inequality (1). – Diego S. Rodrigues Nov 23 '17 at 18:21
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    @Diego S. Rodrigues: use Hermite-Hadamard's inequality on each interval $[k,k+1]$ (with $1 \leq k < n$), then sum the inequalities you get. – D. Thomine Nov 23 '17 at 23:49
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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 $$ H_n-\log(n)\le\gamma+\frac1{2n-1}\tag4 $$

robjohn
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4

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.

Diego S. Rodrigues
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