If $a_n = \frac{e^{n}}{e^{2n}-1}$ how do I show that $a_{n+1} \leq a_n$? 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.
 A: 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:
$$\frac{a_{n+1}}{a_n}=\frac{e^{n+1}(e^{2n}-1)}{e^n(e^{2n+2}-1)}\leq e\frac{e^{2n}-1}{e^{2n+2}-e^2}=\frac1e<1$$
A: 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 $1.$
A: 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.
A: 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.  
A: 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 $n$.
A: 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.
A: $$\dfrac{d}{dn}a_n=-\coth(n)\cdot a_n\le 0$$
So $a_{n+r}\le a_n$ and this is true, in particular, also for $r=1$.
A: 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.
A: $e^{2n+1} - \frac{1}{e} - (e^{2n} - 1)\geq 0$
$e^{2n+1} - \frac{1}{e} \geq e^{2n} - 1$
$\frac{1}{e^{2n} - 1} \geq \frac{1}{e^{2n+1} - \frac{1}{e}}$
$\frac{e^n}{e^{2n} - 1} \geq \frac{e^{n}}{e^{2n+1} - \frac{1}{e}}$
$\frac{e^n}{e^{2n} - 1} \geq \frac{e^{n+1}}{e^{2n+2} - 1}$.
