$\frac{1}{n}+\frac{1}{n+1}+\cdots+\frac{1}{2n}>\frac{2}{3}$ This is from the book Problems in Mathematical Analysis I by Kaczor and Nowak:
Show that, for $n\in \mathbb{N}$, 
$$\frac{1}{n}+\frac{1}{n+1}+\cdots+\frac{1}{2n}>\frac{2}{3}$$
The solution in the back of the book says to apply the Arithmetic Harmonic mean inequality, but when I try to do so I get that 
$$\frac{\sum_{i=0}^n (n+i)}{n+1}\ge \frac{1}{\sum_{i=0}^n\frac{1}{n+i}}$$
and the left hand side is equal to 
$$\frac{n(n+1)+\frac{n(n+1)}{2}}{n+1}=\frac{3n}{2}$$
so it looks like the best I can do is 
$$\sum_{i=0}^n\frac{1}{n+i}\ge \frac{2}{3n}$$
what am I missing here?
 A: As pointed by r9m, Jensen's inequality is enough, since $\frac{1}{x}$ is a convex function on $\mathbb{R}^+$, hence:
$$\sum_{k=n}^{2n}\frac{1}{k}\geq \frac{1}{n}+n\left(\frac{1}{n}\sum_{k=n+1}^{2n} k\right)^{-1}=\frac{1}{n}+\frac{2n}{3n+1}\geq\frac{2}{3}.$$
As an alternative,
$$ \frac{1}{k}\geq \log(k+1)-\log k+\frac{1}{2}\left(\frac{1}{k}-\frac{1}{k+1}\right) $$
implies:
$$ \sum_{k=n}^{2n}\frac{1}{k}\geq \log\left(\frac{2n+1}{n}\right)+\frac{n+1}{2n(2n+1)}.$$
A: Since $f(x)=\frac{1}{x}$ is convex, Jensen's inequality gives
$$
\begin{align}
\frac1{n+1}\sum_{k=n}^{2n}\frac1k
&\ge\left(\frac1{n+1}\sum_{k=n}^{2n}k\right)^{-1}\\
&=\frac{2(n+1)}{(4n^2+2n)-(n^2-n)}\\
&=\frac{2(n+1)}{3n(n+1)}
\end{align}
$$
Therefore,
$$
\bbox[5px,border:2px solid #C0A000]{\sum_{k=n}^{2n}\frac1k
\ge\frac23 \frac{n+1}{n}}
$$
A: $$\frac{\frac1n+\frac1{n+1}+\ldots+\frac1{2n}}{n+1}\ge\frac{n+1}{n+(n+1)+\ldots+(2n)}=\frac{n+1}{n(n+1)+\frac{n(n+1)}2}\implies$$
$$\frac1n+\frac1{n+1}+\ldots+\frac1{2n}\ge\frac{(n+1)^2}{\frac32n(n+1)}=\frac23\frac{(n+1)^2}{n^2+n}\ge\frac23$$
A: $$\sum_{k=n}^{2n}{1\over k} \ge \int_n^{2n}{dt\over t} = \ln(2).$$
