limit of $a_n = \left( \frac{ \sqrt{k} - 1 }{ \sqrt{k} + 1} \right )^n, \quad k\in \mathbb{N}$

Question

How to find the limit of the following sequence:

$a_n = \left( \frac{ \sqrt{k} - 1 }{ \sqrt{k} + 1} \right )^n, \quad \text{for a constant } k\in \mathbb{N}$

$\begin{align} a_n = \left( \frac{ \sqrt{k} - 1 }{ \sqrt{k} + 1} \right )^n &= a_n = \left( \frac{ (\sqrt{k} - 1) (\sqrt{ k } - 1) }{ (\sqrt{k} + 1) (\sqrt{k} - 1) } \right )^n \\ &= a_n = \frac{ (\sqrt{k} - 1 )^{2n}}{ (k - 1)^n} \end{align}$

I do not feel that it did help very much.

Question: How can I go on ?

Answer 1

Hint:

Notice that for $k>1$, we have $0<\frac{\sqrt{k}-1}{\sqrt{k}+1}<1$ and it is independent of $n$

Possible keyword: Geometric sequence.