Proof for $\log\frac{2n+1}{n+1}<\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n}<\log 2$ How can I prove that
$\displaystyle \log\frac{2n+1}{n+1}<\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n}<\log2$
using $\displaystyle \frac{x}{1+x}<\log(1+x)<x$
 A: We have
$$\frac{1}{n+k}>\log\left(1+\frac{1}{n+k}\right) $$
hence:
$$\sum_{k=1}^{n}\frac{1}{n+k}>\log\prod_{k=1}^{n}\frac{n+k+1}{n+k}=\log\frac{2n+1}{n+1}$$
while:
$$\frac{1}{n+k}=\frac{\frac{1}{n+k-1}}{1+\frac{1}{n+k-1}}<\log\left(1+\frac{1}{n+k-1}\right)$$
gives:
$$\sum_{k=1}^{n}\frac{1}{n+k}<\log\prod_{k=1}^{n}\frac{n+k}{n+k-1}=\log2$$
as wanted.
A: Riemann sums do this pretty quickly. $\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n}$ is a 
lower Riemann sum for $\int_n^{2n} {1 \over x}\,dx = \log 2$, and an upper Riemann sum for 
$\int_{n+1}^{2n + 1} {1 \over x}\,dx = \log {2n +1 \over n + 1}$, so one has
$$\log {2n +1 \over n + 1} < \frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n} < \log 2 $$
A: As shown in $(1)$ here,
$$
\begin{align}
H(2n)-H(n)
&=\sum_{k=n+1}^{2n}\frac1k\\
&=\sum_{k=1}^{2n}\color{#C00000}{\frac1k}-\sum_{k=1}^n\color{#00A000}{\frac1k}\\
&=\sum_{k=1}^n\left(\color{#C00000}{\frac1{2k-1}+\frac1{2k}}-\color{#00A000}{\frac1k}\right)\\
&=\sum_{k=1}^n\left(\frac1{2k-1}-\frac1{2k}\right)\\
&=\sum_{k=1}^{2n}(-1)^{k-1}\frac1k\tag{1}
\end{align}$$
which is a series of positive terms, $\left(\frac1{2k-1}-\frac1{2k}\right)$, whose limit is $\log(2)=\sum\limits_{k=1}^\infty(-1)^{k-1}\frac1k$. Thus, each partial sum is less than $\log(2)$. That is,
$$
\sum_{k=n+1}^{2n}\frac1k\lt\log(2)\tag{2}
$$
Furthermore,
$$
\begin{align}
e^{\frac1{n+1}+\frac1{n+2}+\dots+\frac1{2n}}
&\gt\left(1+\frac1{n+1}\right)\left(1+\frac1{n+2}\right)\cdots\left(1+\frac1{2n}\right)\\
&=\frac{2n+1}{n+1}\tag{3}
\end{align}
$$
Taking the log of $(3)$, we get
$$
\sum_{k=n+1}^{2n}\frac1k\gt\log\left(\frac{2n+1}{n+1}\right)\tag{4}
$$
Putting together $(2)$ and $(4)$ yields
$$
\log\left(\frac{2n+1}{n+1}\right)\lt\sum_{k=n+1}^{2n}\frac1k\lt\log(2)\tag{5}
$$
