I can't, I know how to do some by sandwich principle but I couldn't get either of these to work.
$\displaystyle \sqrt[n]{a}\xrightarrow [n\to\infty]{}1\,,\,\,\forall a>0$ , so just check that $\displaystyle 1\leq\sqrt[n]{n+1}\leq \sqrt[n]{2n}=\sqrt[n]{2}\,\sqrt[n]{n}$ ...