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.
 A: 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 $$S=\frac {\text{number of terms}}{2} (\text{first term + last term}) $$
A: Yes, it's correct.
Your sum is an arithmetic sequence.
To sum them up, you can use
$$\sum_{i=1}^m a_i = \frac{m}2 (a_1 + a_m) $$
Here we have $m+1$ terms, the first term is $n$ and the last term if $2n$.
$$\sum_{i=n}^{2n} i  = \frac{(n+1)}2 (n + 2n)=\frac{3n}{2}(n+1)$$
We have
$$\frac23 \cdot \frac{n+1}{n} = \frac23 + \frac2{3n} > \frac23$$
A: For the original problem
$$S_n=\sum_{k=0}^n \frac 1{n+k}$$ you also could using harmonic numbers since
$$S_n=\sum_{k=0}^n \frac 1{n+k}=\sum_{k=1}^{2n} \frac 1{k}-\sum_{k=1}^{n-1}\frac 1{k}= H_{2 n}-H_{n-1}$$ Now, the asymptotics of
$$H_p=\log (p)+\gamma +\frac{1}{2 p}-\frac{1}{12p^2}+O\left(\frac{1}{p^4}\right)$$ Applying twice and using long division to make it simpler
$$S_n=\log (2)+\frac{3}{4 n}+\frac{1}{16 n^2}+O\left(\frac{1}{n^4}\right)$$ and tis very few terms are always larger than $\frac 23$
