How to show the sequence is monotone

Question

"$u_n = \frac{2}{1+e^{-n}}$. Show that $u_n$ is monotone."

My approach would be to consider |$u_{n+1} - u_n$| = |$\frac{2}{1+e^{-n-1}} - \frac{2}{1+e^{-n}}$|.

However I'm not sure the best way to simpify this to show that |$u_{n+1} - u_n$| $>0$. For instance, I tried cross multiplying but then I couldn't manage to simplify it any further.

Another approach I considered was that because the numerators are the same, we can just compare the denominators instead. Then we can get it down to $e^{-n-1} - e^{-n}$, but I'm not sure where I'd go from here.

(note that this is a past exam question, where we aren't allowed to use calculators).

Answer 1

Using functional approach you can write $u_n = \dfrac{2e^n}{1+e^n}=f(e^n)\implies f(x) = \dfrac{2x}{1+x}\implies f'(x) = \dfrac{2}{(1+x)^2} > 0$. Thus $u_n$ is monotonically increasing.

Answer 2

As $e^{-n} > e^{-n-1}$ we get $$ 1 + e^{-n} > 1 + e^{-n-1} $$ and thus $$ \frac{2}{1 + e^{-n}} < \frac{2}{1 + e^{-n-1}}. $$