Show that $e < \big( \frac{2n + 1}{2n - 1} \big)^n$, for $n \in \mathbb{Z}^{+}$

Question

Using the fact that $e^x - 1 < \frac{2x}{2 - x}$ for $0 < x < 2$, prove that $e < \big (\frac{2n + 1}{2n - 1} \big)^n$, for $n \in \mathbb{Z}^{+}$

My line of thought went like this:

Putting $x = 1$ in the intitial statement we have $e < 3$. I calculated a few values of $\big(\frac{2n + 1}{2n - 1} \big)^n$ and it seems to be a decreasing function. Additionally $\lim_{n \rightarrow \infty} \big(\frac{2n + 1}{2n - 1} \big)^n = e $. If it is true that the function is decreasing, then that means that for $n \in [1, \infty)$, the function lies between $3$ and $e$, which implies that it is greater than $e$.

The thing is, I don't have an easy way to prove that the function is decreasing. Its derivative is very messy. Is there an easy way to do this, or a better way to solve the problem?

Answer 1

The given inequality is equivalent to $$ 2n\,\text{arctanh}\left(\tfrac{1}{2n}\right) > 1 $$ which is trivial since $\text{arctanh}(x)$ is increasing and convex on $(0,1)$: $$ \text{arctanh}(x) = x +\frac{x^3}{3}+\frac{x^5}{5}+\frac{x^7}{7}+\ldots $$

Answer 2

Assume:

$ e^x \lt 1+\dfrac{2x}{2-x} = \dfrac{2+x}{2-x}$ , $0<x<2.$

$x:= 1/n $, $ n \in \mathbb{Z^+} $.

$e^{1/n} \lt \dfrac{2+1/n}{2-1/n} = \dfrac{2n+1}{2n-1}.$

$f(x)= x^n$ is monotonically increasing for $x >0.$

Hence :

$e \lt \Big (\dfrac{2n+1}{2n-1} \Big )^n$ for $n\in \mathbb{Z^+}.$

Note:

Concerning the assumption:

$f(x):= \dfrac{2x}{2-x} +1 - e^x .$

$f(x) =0.$

$f'(x) = \dfrac{4}{(2-x)^2} -e^x.$

$f'(x) >0$ for $0<x<2.$

Hence $f(x) >0$ in the above range .

Answer 3

Let's use that:

$$\big (1+\frac{1}{k} \big)^{k}<e<\big (1+\frac{1}{k} \big)^{k+1}<\big (1+\frac{2}{k} \big)^{k+1}$$

thus:

$$\big (\frac{2n + 1}{2n - 1} \big)^{2n}=\big (1+\frac{2}{2n - 1} \big)^{2n}=\big (1+\frac{2}{k} \big)^{k+1}>e \quad \square$$