Inequality $\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{3n+1}>1$

Question

Show that $$\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{3n+1}>1,\:\forall n\in\mathbb{N}$$ This is a 9th grade problem. I was trying to take the greatest numerator, which is the last numerator of the last fraction. But there are only $2n+1$ terms. Right? After that I have no idea. Thx!

Answer 1

By C-S $$\sum_{k=1}^{2n+1}\frac{1}{n+k}\geq\frac{(2n+1)^2}{\sum\limits_{k=1}^{2n+1}(n+k)}=\frac{(2n+1)^2}{\frac{(2(n+1)+2n)(2n+1)}{2}}=1.$$

Answer 2

There are $2n+1$ terms in the sum, you just need to pair up the terms symmetrically from both ends, take average and compare with the term in the middle. $$\begin{align}\sum_{k=n+1}^{3n+1} \frac{1}{k} &= \sum_{k=-n}^n\frac{1}{2n+1+k} = \frac12\sum_{k=-n}^n\left(\frac{1}{2n+1+k} + \frac{1}{2n+1-k}\right)\\ &= \sum_{k=-n}^n \frac{2n+1}{(2n+1)^2-k^2} \stackrel{\color{blue}{\text{ assume } n> 0}}{>} \sum_{k=-n}^n \frac{1}{2n+1} = \frac{2n+1}{2n+1} = 1\end{align}$$

Answer 3

Use the method given in this answer by Jack D'Aurizio.

Note: $H_n=1+\frac12+\frac13+\cdots+\frac1n=\sum_{k=1}^n \frac1n$ is called $n$-th harmonic number.

Then: $$\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{3n+1}=\sum_{k=1}^{2n+1}\frac{1}{n+k}=H_{3n+1}-H_{n}.$$ Consider the sequence: $a_n=H_{3n+1}-H_n$. We will show that it is an increasing sequence: $$a_{n+1}-a_n=(H_{3(n+1)+1}-H_{n+1})-(H_{3n+1}-H_n)=\\ (H_{3n+4}-H_{3n+1})-(H_{n+1}-H_n)=\\ \frac{1}{3n+4}+\frac{1}{3n+3}+\frac{1}{3n+2}+\frac{1}{3n+1}-\frac{1}{n+1}>\\ \frac{1}{3n+4}+\frac{1}{3n+3}+\frac{1}{3n+\color{red}{3}}+\frac{1}{3n+\color{red}{3}}-\frac{1}{n+1}=\\ \frac{1}{3n+4}>0 \Rightarrow a_{n+1}>a_n.$$ Hence: $$a_1=H_{3+1}-H_1=\sum_{k=1}^{2+1} \frac{1}{1+k}=\frac12+\frac13+\frac14=\frac {13}{12}>1.$$