TITLE: $\frac{1}{n}+\frac{1}{n+1}+\cdots+\frac{1}{2n}>\frac{2}{3}$
QUESTION [5 upvotes]: 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?

REPLY [3 votes]: 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,
$$
\bbox[5px,border:2px solid #C0A000]{\sum_{k=n}^{2n}\frac1k
\ge\frac23 \frac{n+1}{n}}
$$