Inequality with $\frac{1}{n}+\frac{1}{n+1}+\cdots+\frac{1}{2n}$

Question

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 $(1)$?

Answer 1

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.

Answer 2

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:

$$1-\frac12+\frac13-\cdots+\frac1{2n+1}\leq \frac{1}{2n+1}+\cdots+\frac{1}{4n+2}$$