Show that, for all $n > 1: \log \frac{2n + 1}{n} < \frac1n + \frac{1}{n + 1} + \cdots + \frac{1}{2n} < \log \frac{2n}{n - 1}$ I'm learning calculus, specifically limit of sequences and derivatives, and need help with the following exercise:

Show that for every $n > 1$,
$$\log \frac{2n + 1}{n} < \frac1n + \frac{1}{n + 1} + \cdots + \frac{1}{2n} < \log \frac{2n}{n - 1} \quad \quad (1)$$

Important: this exercise is the continuation of a previous problem showing that, for every $n > 1$,
$$\frac{1}{n + 1} < \log(1 + \frac1n) < \frac1n \quad \quad (2)$$
A detailed solution of the latter inequality using the MVT can be found here.
Now back to inequality $(1)$. My first guess was to use mathematical induction in order to prove it but I didn't get far. I think I should make use of inequality $(2)$ from the previous exercise but I'm stuck here.
 A: Hint:
To prove the left half of the inequality, start by noting that
$$
\begin{align*}
\frac{1}{n}+\frac{1}{n+1}+\cdots+\frac{1}{2n}&>\log\left(1+\frac{1}{n}\right)+\log\left(1+\frac{1}{n+1}\right)+\cdots\log\left(1+\frac{1}{2n}\right)\\
&=\log\left(\frac{n+1}{n}\right)+\log\left(\frac{n+2}{n+1}\right)+\cdots+\log\left(\frac{2n+1}{2n}\right).
\end{align*}
$$
How can you simplify this sum of logarithms?
A: For the LHS it is $$\frac { 1 }{ n } +\frac { 1 }{ n+1 } +\cdots +\frac { 1 }{ 2n } =\int _{ n }^{ n+1 }{ \frac { 1 }{ n } dx } +\int _{ n+1 }^{ n+2 }{ \frac { 1 }{ n+2 } dx } +...+\int _{ 2n }^{ 2n+1 }{ \frac { 1 }{ 2n } dx } >\\ >\int _{ n }^{ n+1 }{ \frac { 1 }{ x } dx } +\int _{ n+1 }^{ n+2 }{ \frac { 1 }{ x } dx } +...+\int _{ 2n }^{ 2n+1 }{ \frac { 1 }{ x } dx } =\int _{ n }^{ 2n+1 }{ \frac { 1 }{ x }  } dx=\log  \frac { 2n+1 }{ n }  $$
and for the RHS it is $$\frac { 1 }{ n } +\frac { 1 }{ n+1 } +\cdots +\frac { 1 }{ 2n } =\int _{ n }^{ n+1 }{ \frac { 1 }{ n } dx } +\int _{ n+1 }^{ n+2 }{ \frac { 1 }{ n+2 } dx } +...+\int _{ 2n }^{ 2n+1 }{ \frac { 1 }{ 2n } dx } <\\ <\int _{ n-1 }^{ n }{ \frac { 1 }{ x } dx } +\int _{ n }^{ n+1 }{ \frac { 1 }{ x } dx } +...+\int _{ 2n-1 }^{ 2n }{ \frac { 1 }{ x } dx } =\int _{ 2n-1 }^{ 2n }{ \frac { 1 }{ x }  } dx=\log  \frac { 2n }{ n-1 } \\ $$
A: For the sake of completeness, let we show how to get a tight upper bound with the same technique used by Nick Peterson for the lower bound.
Since for any $|x|<1$ we have
$$ \text{arctanh}(x) = \frac{1}{2}\log\left(\frac{1+x}{1-x}\right) = x + \frac{x^3}{3}+\frac{x^5}{5}+\ldots \tag{1} $$
by evaluating $(1)$ at $x=\frac{1}{n}$ we get:
$$ \sum_{k=n}^{2n}\frac{1}{k} < \frac{1}{2}\sum_{k=n}^{2n}\log\left(\frac{k+1}{k-1}\right)=\frac{1}{2}\log\prod_{k=n}^{2n}\frac{k+1}{k-1}=\frac{1}{2}\log\left(\frac{4n+2}{n-1}\right). \tag{2}$$
