Inequality $\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{3n+1}>1$ 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!
 A: 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.
$$\begin{align}\sum_{k=n+1}^{3n+1} \frac{1}{k} 
&= \sum_{k=-n}^n\frac{1}{2n+1+k}
= \frac12\sum_{k=-n}^n\left(\frac{1}{2n+1+k} + \frac{1}{2n+1-k}\right)\\
&= \sum_{k=-n}^n \frac{2n+1}{(2n+1)^2-k^2}
\stackrel{\color{blue}{\text{ assume } n> 0}}{>} \sum_{k=-n}^n \frac{1}{2n+1} = \frac{2n+1}{2n+1} = 1\end{align}$$
A: By C-S
$$\sum_{k=1}^{2n+1}\frac{1}{n+k}\geq\frac{(2n+1)^2}{\sum\limits_{k=1}^{2n+1}(n+k)}=\frac{(2n+1)^2}{\frac{(2(n+1)+2n)(2n+1)}{2}}=1.$$
A: 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:
$$a_1=H_{3+1}-H_1=\sum_{k=1}^{2+1} \frac{1}{1+k}=\frac12+\frac13+\frac14=\frac {13}{12}>1.$$
