I have problems with finding: $$\lim_{n\to\infty} \sqrt[n]{{\frac{1}{\sqrt{3}}\Bigg(\Bigg(\frac{1+\sqrt{3}}{2}\Bigg)^n}-\Bigg(\frac{1-\sqrt{3}}{2}\Bigg)^n\Bigg)}$$ I tried to do it following way:
$\displaystyle\lim_{n\to\infty} \sqrt[n]{{\frac{1}{\sqrt{3}}\Bigg(\Bigg(\frac{1+\sqrt{3}}{2}\Bigg)^n}-\Bigg(\frac{1-\sqrt{3}}{2}\Bigg)^n\Bigg)}$
$\displaystyle\lim_{n\to\infty} \sqrt[n]{\frac{1}{\sqrt{3}}}\cdot\lim_{n\to\infty} \sqrt[n]{\Bigg(\Bigg(\frac{1+\sqrt{3}}{2}\Bigg)^n-\Bigg(\frac{1-\sqrt{3}}{2}\Bigg)^n\Bigg)}=$
$1\cdot\displaystyle\lim_{n\to\infty} \sqrt[n]{\Bigg(\Bigg(\frac{1+\sqrt{3}}{2}\Bigg)^n-\Bigg(\frac{1-\sqrt{3}}{2}\Bigg)^n\Bigg)}$
Now I used formula for difference of powers: $$$$
$\Big(\frac{1+\sqrt{3}}{2}\Big)^n-\Big(\frac{1-\sqrt{3}}{2}\Big)^n=\Big(\frac{1+\sqrt{3}}{2}-\frac{1-\sqrt{3}}{2}\Big)\cdot\Big( \Big(\frac{1+\sqrt{3}}{2}\Big)^{n-2}\Big(\frac{1+\sqrt{3}}{2}\Big)+\Big(\frac{1+\sqrt{3}}{2}\Big)^{n-3}\Big(\frac{1+\sqrt{3}}{2}\Big)\Big(\frac{1-\sqrt{3}}{2}\Big)+\Big(\frac{1+\sqrt{3}}{2}\Big)^{n-4}\Big(\frac{1+\sqrt{3}}{2}\Big)\Big(\frac{1-\sqrt{3}}{2}\Big)+...+\Big(\frac{1-\sqrt{3}}{2}\Big)^{n-1}\Big)$
In the last parenthesis, I saw two geometric series and I tried to add them, however, it quickly appeared that there will be other geometric series and here is where I am a bit helpless (it is getting very nasty very quickly). Do you have any hints to move it in maybe another way? I would be very grateful, thanks!
For the beginning prove the following facts $$ \lim\limits_{n\to\infty} \sqrt[n]{a}=1 \quad\text{ for }\quad a>0 $$ $$ \lim\limits_{n\to\infty} \sqrt[n]{x^n-y^n}=x\quad\text{ for }\quad x>|y| $$ then apply them to the limit $$ \lim\limits_{n\to\infty}\sqrt[n]{a(x^n-y^n)} $$