<s>[INST] AM>HM Problem $\frac{1}{n+1}+...+\frac{1}{3n+1}>1$ I am having difficulty solving one of the problems from "Problems in Mathematical Analysis I" -  W. J. Kaczor;M. T. Nowak .
It's a problem 1.2.5 b), and it goes like this:

1.2.5. For $n \in \mathbb{N}$, verify the following claims: 
$$\tag{b} \qquad \dfrac{1}{n + 1} + \dfrac{1}{n + 2} + \dfrac{1}{n + 3} + \ldots + \dfrac{1}{3n + 1} \, > \, 1$$

In solutions it says: "Use the arithmetic-harmonic mean inequality!"
I tried to apply it on whole inequality but got:
\begin{align}
& \dfrac{\frac{1}{n+1}+\ldots+\frac{1}{3n+1}}{2n}\, >\, \dfrac{2n}{n+1+\ldots+3n+1} \\ 
\implies & \frac{1}{n+1}+\ldots+\frac{1}{3n+1}>\frac{8n^2}{2n(n+1+3n+1)} \\
\implies & \frac{1}{n+1}+\ldots+\frac{1}{3n+1}>\frac{2n}{2n+1}
\end{align}
 [/INST] You claim that $(n+1)+\cdots+(3n+1) = 2n((n+1)+(3n+1))$, which is not correct. 
First of all, there are $2n+1$ terms, not $2n$ (as user8734617 says in a comment).
Also, you need to multiply with the average of the first and the last term. 
So you actually get $$\frac{(2n+1)^2} { (2n+1) \left( \frac{(n+1)+(3n+1)}{2} \right)} = \frac{(2n+1)^2}{(2n+1)^2} = 1$$

For an alternative solution, try induction. 
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