Limit $\lim\limits_{n\to\infty} \sqrt[n]{\frac1{\sqrt3}\left(\left(\frac{1+\sqrt3}2\right)^n-\left(\frac{1-\sqrt3}2\right)^n\right)}$ 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!

Let $p = \frac{1+\sqrt{3}}2$ and $q = \frac{1-\sqrt{3}}2$ and denote your sequence as $$a_n = \frac{p^n-q^n}{\sqrt3}.$$
We have
$$\frac{a_{n+1}}{a_n} = \frac{p^{n+1}-q^{n+1}}{p^n-q^n} = \frac{p-q\left(\frac{p}{q}\right)^n}{1-\left(\frac{p}{q}\right)^n} \xrightarrow{n\to\infty} p$$
so by this question it also follows that
$$\lim_{n\to\infty} \sqrt[n]{a_n} = p$$
because if the ratio $\left(\frac{a_{n+1}}{a_n}\right)_n$ converges, then the $n$-th root $\left(\sqrt[n]{a_n}\right)_n$ also converges and to the same limit.
