Prove that $\frac{1}{n}+\frac{1}{n+1}+...+\frac{1}{2n}>\frac{2}{3}$ Prove that $\frac{1}{n}+\frac{1}{n+1}+...+\frac{1}{2n}>\frac{2}{3}$ for any $n \in N$
I used AM $\geq$ HM and got $$\frac{1}{n}+\frac{1}{n+1}+...+\frac{1}{2n}\geq\frac{(n+1)^2}{n+n+1+...+2n}=\frac{(n+1)^2}{\frac{3}{2}n(n+1)}>\frac{2}{3}$$
My question is whether this solution is correct and also why $n+n+1+...+2n = \frac{3}{2}n(n+1)$, since I used wolfram to compute that and cannot see it myself.
Any tips on how to compute sums like this or other possible solutions for this problem would be appreciated as well.

That is correct, and $$n+(n+1) +(n+2) +\dots +(n+n) \\ =(n+1)\times n +(1+2+3+\dots +n) \\ = n(n+1) + \frac{n(n+1)}{2} \\ =\frac 32 n(n+1) $$
Alternatively, you can use the formula for the sum of an arithmetic  progression $$S=\frac {\text{number of terms}}{2} (\text{first term + last term}) $$
