If $a_n = \frac{e^{n}}{e^{2n}-1}$ how do I show that $a_{n+1} \leq a_n$?

Question

Let $$a_n = \frac{e^{n}}{e^{2n}-1}$$ How do I show that $a_{n+1} \leq a_n$?

I don't know how to deal with the $-1$ in the denominator.

Answer 1

Note that $1/a_n=e^n-e^{-n}=2\sinh n$. Since the function $\sinh$ is increasing, so is the sequence $1/a_n$, and therefore, $a_n$ is decreasing.

Another way:

$$\frac{a_{n+1}}{a_n}=\frac{e^{n+1}(e^{2n}-1)}{e^n(e^{2n+2}-1)}\leq e\frac{e^{2n}-1}{e^{2n+2}-e^2}=\frac1e<1$$

Answer 2

Assuming $n \gt 0$ throughout:

$a_n = \dfrac{e^{n}}{e^{2n}-1} =\dfrac{1}{e^{n}-e^{-n}}$.

Since $e \gt 1$, you have $e^{n}$ is an increasing function of $n$ and is greater than $1$,

so $e^{-n} = \dfrac{1}{e^n}$ is a decreasing function of $n$ and is less than $1$,

so $e^{n}-e^{-n}$ is a positive and increasing function of $n$,

so $a_n = \dfrac{1}{e^{n}-e^{-n}}$ is a positive and decreasing function of $n$.

Answer 3

Need to show:

$$\frac{e^{n+1}}{e^{2n+2}-1} \leq \frac{e^{n}}{e^{2n}-1}$$

that is

$$(e^{2n}-1)e^{n+1} \leq (e^{2n+2}-1)e^{n}$$ $$e^{3n+1}-e^{n+1} \leq e^{3n+2}- e^n$$ $$e^n(e^{2n+2}-e^{2n+1}+e-1) \geq 0$$

Now analysing this shold be easy. Both factors are positive.

Answer 4

The inequality to be proved is $$ \frac{e^{n+1}}{e^{2n+2}-1}\le\frac{e^n}{e^{2n}-1} $$ Write $x=e^n$ and note that, for $n>0$, $e^n>1$, so $x>1$ and $x^2>1$. Then the inequality to be proved is $$ \frac{ex}{e^2x^2-1}\le\frac{x}{x^2-1} $$ that is equivalent to $$ ex(x^2-1)\le x(e^2x^2-1) $$ Simplify by $x$, getting the equivalent inequality $$ ex^2-e\le e^2x^2-1 $$ or, again equivalently, $$ 1-e\le x^2(e^2-e) $$ The left hand side is negative, while the right hand side is positive, so the inequality is true.

Answer 5

$$\dfrac{d}{dn}a_n=-\coth(n)\cdot a_n\le 0$$

So $a_{n+r}\le a_n$ and this is true, in particular, also for $r=1$.

Answer 6

I would run with

$$\begin{align} a_n-a_{n+1}&=\frac{e^n}{e^{2n}-1}-\frac{e^{n+1}}{e^{2n+2}-1} \\&=\frac{e^n(e^{2n+2}-1)-e^{n+1}(e^{2n}-1)}{(e^{2n}-1)(e^{2n+2}-1)} \\&=\frac{e^2e^{3n}-e^n-ee^{3n}+ee^n}{(e^{2n}-1)(e^{2n+2}-1)} \\&=\frac{e^{3n}(e^2-e)+e^n(e-1)}{(e^{2n}-1)(e^{2n+2}-1)}\geq0. \end{align}$$

So...

Alternatively, define $f:\mathbb{R}\rightarrow \mathbb{R}$ by $x\mapsto f(x)$ where

$$f(x)=\frac{e^x}{e^{2x}-1}.$$ Now using the quotient rule we have

$$f'(x)=-\frac{e^{3x}+e^{x}}{(e^{2x}-1)^2}<0,$$ so that $f$ is decreasing. Restricting to $x=n\in\mathbb{N}$ doesn't change this fact: $a_n=f(n)$ is decreasing.

Answer 7

Look at the ratio:$$\frac{a_{n+1}}{a_n} = \frac{e^{n+1}}{e^{2n+2}-1}\frac{e^{2n}-1}{e^n}=\frac{e^{2n+1}-e}{e^{2n+2}-1}.$$Since $e^{2n+1}<e^{2n+2}$ and $e>1,$ the numerator is less than the denominator. So this ratio is less than $1.$

Answer 8

here is one way to see $\frac{e^n}{e^{2n} - 1}$ is decreasing with $n.$ it is enough to show that $\frac{x}{x^2-1}$ is decreasing for $x > 1.$ we will use the simple facts $\frac1{x}$ and its translates are decreasing on $(1, \infty).$ we have $$\frac{x}{x^2-1} = \frac12\left(\frac1{x-1}+\frac1{x+1}\right)$$ is the average of two decreasing functions, therefore is decreasing.

Answer 9

$e^{2n+1} - \frac{1}{e} - (e^{2n} - 1)\geq 0$

$e^{2n+1} - \frac{1}{e} \geq e^{2n} - 1$

$\frac{1}{e^{2n} - 1} \geq \frac{1}{e^{2n+1} - \frac{1}{e}}$

$\frac{e^n}{e^{2n} - 1} \geq \frac{e^{n}}{e^{2n+1} - \frac{1}{e}}$

$\frac{e^n}{e^{2n} - 1} \geq \frac{e^{n+1}}{e^{2n+2} - 1}$.