How to prove a sequence is increasing

Question

The overall problem is given the recursive sequence define as $s_0=\sqrt[4]{2}$ and $s_{k+1}=\sqrt[4]{2+s_k}$ and proving $\lim_{k \to \infty}s_k$ exists. So my first course of action was to prove the sequence was bounded which I did just fine (I believe). Now I need to show the sequence is increasing. I am trying to achieve $s_k<s_{k+1}$ but algebraically I can not seem to achieve this. I have tired using $s_{k+1}<s_{k+2}$ to give me little more to work with but I cannot complete the proof.

Answer 1

For $x > 0$, consider: $f(x) = \sqrt[4]{x+2}\implies f'(x) = \dfrac{1}{4\sqrt[4]{(x+2)^3}}>0$. This means: $s_1 > s_0 \implies f(s_1) > f(s_0)\implies s_2 > s_1\implies f(s_2) > f(s_1)\implies s_3 > s_2$, and by induction $s_n > s_{n-1}$: increasingness follows.

Answer 2

$s_{k+1}^4-s_k^4=s_{k}-s_{k-1}\implies s_{k+1}-s_k>0$ if $s_{k}-s_{k-1}>0$, since $b^4-a^4=(b^2-a^2)(b^2+a^2)=(b-a)(b+a)(b^2+a^2)$.