$ \frac{1}{2} + \dots + \frac{1}{n} \le \log n $

Question

could anyone give me any hint how to prove this ?

$$ \frac{1}{2} + \dots + \frac{1}{n} \le \log n $$

just came acroos this expression in my book.

Answer 1

Fix a positive integer $k$. Using the fact that $g(x) = x^{-1}$ is decreasing on $[k,k+1]$, we have $$(k+1)^{-1} \leq x^{-1} \leq k^{-1} \text{ for } x \in [k,k+1]$$ Integrating over $[k,k+1]$ yields $$\int_k^{k+1} (k+1)^{-1} dx\leq\int_k^{k+1} x^{-1} dx\leq\int_k^{k+1} k^{-1} dx$$ $$(k+1)^{-1} \leq\log(k+1) - \log(k)\leq k^{-1} $$ Summing up over $k = 1,2, \ldots, n-1$ we have $$\sum_{k=1}^{n-1} (k+1)^{-1} \leq \sum_{k=1}^{n-1} [\log(k+1) - \log(k)] \leq \sum_{k=1}^{n-1} k^{-1} $$ $$2^{-1} + 3^{-1} + \ldots + n^{-1}\leq \log(n) \leq 1 + 2^{-1} + \ldots + (n-1)^{-1} $$

Answer 2

Note that $$\log n=\int_1^n \frac{1}{x} \, dx$$

Note that if we take right handed sums with subinterval 1, we see that the sum of them is $\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}$. And because the right handed sums are lower than the graph of $f(x)=\frac1x$ , we see that their area is smaller than the actual integral, that is, $\log x$.

Answer 3

Use the inequality $$\frac{1}{n+1} \leqslant \ln \left( 1+\frac{1}{n} \right)$$ to prove yours by induction.