TAOCP 1.2.7 Exercise 4
Statement 1 is false, since $H_1=1>\ln 1=0$.
Section 1.2.7: Harmonic Numbers
Exercise 4. [10] Decide which of the following statements are true for all positive integers $n$:
- $H_n < \ln n$.
- $H_n > \ln n$.
- $H_n > \ln n + \gamma$.
Verified: no
Solve time: -
Statement 1 is false, since $H_1=1>\ln 1=0$.
Statement 2 is true for all positive integers $n$. Since $1/x$ is decreasing on $[1,n]$,
$$ \int_k^{k+1}\frac{dx}{x}<\frac{1}{k} \qquad (1\le k\le n-1). $$
Summing these inequalities gives
$$ \int_1^n\frac{dx}{x} < \sum_{k=1}^{n-1}\frac{1}{k} = H_n-\frac1n < H_n. $$
Therefore
$$ \ln n<H_n. $$
Statement 3 is false for all positive integers $n$. By the definition of Euler's constant,
$$ \gamma=\lim_{m\to\infty}(H_m-\ln m). $$
Hence $H_n-\ln n$ approaches $\gamma$ as $n\to\infty$. If $H_n>\ln n+\gamma$ were true for every $n$, then
$$ H_n-\ln n>\gamma $$
for every $n$, contradicting the fact that the sequence converges to $\gamma$. Therefore statement 3 cannot hold for all $n$.
Thus the correct classification is
$$ \boxed{\text{1 false,\quad 2 true,\quad 3 false.}} $$