How to find the infimum

Question

I have a task from my lecturer, but i don't know how to solve it.

Let $k \in [0,1)$, find the infimum of $\sqrt{(1+k^{2n})^2 + (1+k^{2n-1})^2}$ for all $n \in \mathbb{N}$

Answer 1

I would first show that $k^n$ is decreasing when $k\in[0,1)$.

Suppose $k\in[0,1)$. Then, for any positive number $x>0$, we have $kx <x$. Therefore, $k^{n}$ is a decreasing sequence, and must have a limit $\beta$. As this sequence converges, the difference between the terms converges to zero, and therefore $k^{n+1}-k^{n}\to 0$ which forms $\beta(k-1)=0$, from which we get $\beta = 0$. So, $\lim_{n\to\infty}k^{n}=0$ for $k\in[0,1)$.

It then follows that $\lim_{n\to\infty}k^{2n}=0$ for $k\in[0,1)$ and $\lim_{n\to\infty}k^{2n-1}=0$ for $k\in[0,1)$. Therefore the infimum is

$$\lim_{n\to\infty}\sqrt{(1+k^{2n})^2 + (1+k^{2n-1})^2}=\sqrt{2}$$