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?
Best Answer
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
$$H_n>\log(n)+\frac12$$