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I have problems with finding limit of $$\lim_{n\rightarrow\infty} \sqrt[n]{{\frac{1}{\sqrt{3}}((\frac{1+\sqrt{3}}{2})^n}-(\frac{1-\sqrt{3}}{2})^n)}$$I tried to do it following way:
$\lim_{n\rightarrow\infty} \sqrt[n]{{\frac{1}{\sqrt{3}}((\frac{1+\sqrt{3}}{2})^n}-(\frac{1-\sqrt{3}}{2})^n)}$=
$\lim_{n\rightarrow\infty} \sqrt[n]{\frac{1}{\sqrt{3}}} *\lim_{n\rightarrow\infty} \sqrt[n]{((\frac{1+\sqrt{3}}{2})^n-(\frac{1-\sqrt{3}}{2})^n)}=$
$1*\lim_{n\rightarrow\infty} \sqrt[n]{((\frac{1+\sqrt{3}}{2})^n-(\frac{1-\sqrt{3}}{2})^n)}$
Now i used formula $(\frac{1+\sqrt{3}}{2})^n-(\frac{1-\sqrt{3}}{2})^n=(\frac{1+\sqrt{3}}{2}-\frac{1-\sqrt{3}}{2})*$
$*\left ( (\frac{1+\sqrt{3}}{2})^{n-2}(\frac{1+\sqrt{3}}{2})+(\frac{1+\sqrt{3}}{2})^{n-3}(\frac{1+\sqrt{3}}{2})(\frac{1-\sqrt{3}}{2})+(\frac{1+\sqrt{3}}{2})^{n-4}(\frac{1+\sqrt{3}}{2})(\frac{1-\sqrt{3}}{2})+...+(\frac{1-\sqrt{3}}{2})^{n-1}\right ) $
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 there is where i am a bit helpless (it is getting gvery nasty very quickly). Do you have any hints to move it in maybe other way? I would be very grateful, thanks!

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up vote 4 down vote accepted

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 $$ \lim\limits_{n\to\infty}\sqrt[n]{a(x^n-y^n)} $$

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I feel like an idiot, so simple it was. Nvm, thank you for answer – fdhd Nov 9 '12 at 9:04

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