Inequality with $\frac{1}{n}+\frac{1}{n+1}+\cdots+\frac{1}{2n}$ Inspired by this recent question, I suggest this.
Let $n=2,3,4, \ldots .$ Then
$$
\frac{7}{12} < \cfrac 1 {1 + \cfrac {1^2} {1 + \cfrac {2^2} {\ddots + \cfrac \vdots { 1 + \, {n^2} \,}}}}  \leq \frac{1}{n}+\frac{1}{n+1}+\cdots+\frac{1}{2n} \tag1
$$
Could you prove $(1)$?
 A: Here is a brute force method to prove the first inequality. We will see the calculations for $n=2,3,4,5$ and that will be sufficient to prove for general $n$. Here goes: 
$n=2$
$$\frac{7}{12} \leq \cfrac{1}{1+\cfrac{1^2}{1 + 2^2}} \iff \frac{12}{7} \geq 1+\cfrac{1^2}{1 + 2^2} \iff \frac{5}{7} \geq \cfrac{1^2}{1 + 2^2}$$
which is true. For $n=3$, we'll take the calculations from here to get
$$ \frac{5}{7} \geq \cfrac{1^2}{1 + \cfrac{2^2}{1+3^2}} \iff \frac{7}{5} \leq 1 + \cfrac{2^2}{1+3^2} \iff \frac{2}{5} \leq \cfrac{2^2}{1+3^2} $$ which is again true. Carrying on for $n=4$
$$ \frac{2}{5} \leq \cfrac{2^2}{1+ \cfrac{3^2}{1+4^2}} \iff 10 \geq 1+ \cfrac{3^2}{1+4^2} \iff 1 \geq \cfrac{1}{1+4^2} $$ which is true. Carrying on for $n=5$, we have
$$ 1 \geq \cfrac{1}{1+ \cfrac{4^2}{1+5^2}} \iff 1 \leq 1+ \frac{4^2}{1+5^2}  \iff 0 \leq \frac{4^2}{1+5^2} $$ which is true.
For $n\geq 6$, the equivalent condition will just be 
$$ 0 \leq \cfrac{4^2}{1+ \cfrac{5^2}{1+ \cfrac{6^2}{1+ ...}}} $$
which is of course true. These calculations are not revealing, but I sometimes find them cute. 
A: The middle term is a generalized continued fraction, with $\{a_1,a_2,\ldots\}=\{1,1,2^2,3^2,\ldots\}$ and $\{b_0,b_1,\ldots\}=\{0,1,1,1,\ldots\}$.
The recurrence relation for the numerator and denominator of the convergents gives 
$$\begin{align}
x_n&=\frac{A_n}{B_n}\\
&=\frac{\mbox{[A024167](https://oeis.org/A024167)}}{n!}\\
&=\frac{n!\left(1-1/2+\frac13-\cdots\pm\frac1n\right)}{n!}\\
&=1-1/2+\frac13-\cdots\pm\frac1n\\
&\to\ln2\\
&\approx0.693147\ldots>\frac7{12}
\end{align}$$
As for the right side, 
$$\begin{align}
\frac{1}{n}+\cdots+\frac{1}{2n}&=\left(1+\cdots+\frac{1}{2n}\right)-\left(1+\cdots+\frac{1}{n-1}\right)\\
&\sim\ln(2n)-\ln(n)\\
&=\ln(2)
\end{align}$$
So for large enough $n$, the middle and the right are very close.
But we are asked to show that 
$$\frac{7}{12}<1-\frac12+\frac13-\cdots+(-1)^{n-1}\frac1n\leq \frac{1}{n}+\cdots+\frac{1}{2n}$$
The first inequality is true for $n\geq4$ since $1-\frac12+\frac13-\frac14=\frac7{12}$ and the remaining bits of the sequence are a net positive.
The right side converges to $\ln(2)$ from above, while the middle converges to $\ln(2)$ alternating between above and below. So it suffices to show the second inequality for odd $n$. So we'd continue by trying to show:
$$1-\frac12+\frac13-\cdots+\frac1{2n+1}\leq \frac{1}{2n+1}+\cdots+\frac{1}{4n+2}$$
