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.

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