TITLE: Demonstrate by induction the inequality: $\ln(1+n)\leq\sum_{i=1}^{n}\frac{1}{i}\leq1+\ln(n)$
QUESTION [5 upvotes]: Kind of stuck in this one. I've tried substracting
$$\ln(1+(n+1))\leq\sum_{i=1}^{n}\frac{1}{i}+\frac{1}{n+1}\leq1+\ln(n+1)$$
at the original inequality and applying properties of the logarithms, but then I don't know how to keep going :/

REPLY [3 votes]: We denote $\displaystyle H_n=\sum_{i=1}^{n}\frac{1}{i}.$
Since the function $x\mapsto \frac{1}{x}$ is decreasing, we have
$$H_{n+1}-1=\sum_{i=2}^{n+1}\frac{1}{i}=\sum_{i=1}^{n}\frac{1}{i+1}\leq \sum_{i=1}^{n}\int_i^{i+1}\frac{dx}{x}=\int_1^{n+1}\frac{dx}{x}=\log( n+1)\leq\sum_{i=1}^{n}\frac{1}{i}=H_n, $$
and thus we conclude the result(with change of index in the second inequality).