<s>[INST] 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!
 [/INST] 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.
$$\begin{align}\sum_{k=n+1}^{3n+1} \frac{1}{k} 
&= \sum_{k=-n}^n\frac{1}{2n+1+k}
= \frac12\sum_{k=-n}^n\left(\frac{1}{2n+1+k} + \frac{1}{2n+1-k}\right)\\
&= \sum_{k=-n}^n \frac{2n+1}{(2n+1)^2-k^2}
\stackrel{\color{blue}{\text{ assume } n> 0}}{>} \sum_{k=-n}^n \frac{1}{2n+1} = \frac{2n+1}{2n+1} = 1\end{align}$$
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