# Demonstrate by induction the inequality: $\ln(1+n)\leq\sum_{i=1}^{n}\frac{1}{i}\leq1+\ln(n)$

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 :/

• Don't forget to accept answers to your old questions if any of them satisfies you. Feb 25, 2013 at 18:32
• I suppose you mean for the lower index of the sum to be $1$ on the title. Feb 25, 2013 at 18:36

For the induction step, it is sufficient to show that $$\log\left( 1+\frac{1}{n+1}\right)\leq \frac{1}{n+1}\leq \log\left( 1+\frac{1}{n}\right)=- \log \left( 1-\frac{1}{n+1}\right).$$
Now recall that the graph of $\log(1+x)$ is below the graph of $x$ by concavity of $\log$.
• I don't follow you, why the $=-log(1-\frac{1}{n+1})$ ? Feb 25, 2013 at 18:59
• I rewrote the rhs like that so that it is easy to show the second inequality, like the first one, simply using $\log(1+x)\leq x$ for all $x>-1$. Feb 25, 2013 at 19:00
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).