<s>[INST] Finding $\lim \frac{(2n^{\frac 1n}-1)^n}{n^2}$. I want to find limit of $a_n= \frac{(2n^{\frac 1n}-1)^n}{n^2}$ as $n\to \infty$.
$\displaystyle a_{n} =\frac{\left( 2n^{\frac{1}{n}} -1\right)^{n}}{n^{2}} =\left(\frac{2}{n^{\frac{1}{n}}} -\frac{1}{n^{\frac{2}{n}}}\right)^{n}$
$\displaystyle  \begin{array}{{>{\displaystyle}l}}
\log a_{n} =n\log\left(\frac{2}{n^{\frac{1}{n}}} -\frac{1}{n^{\frac{2}{n}}}\right) =\frac{\log\left( 1+\left(\frac{2}{n^{\frac{1}{n}}} -\frac{1}{n^{\frac{2}{n}}} -1\right)\right)}{\left(\frac{2}{n^{\frac{1}{n}}} -\frac{1}{n^{\frac{2}{n}}} -1\right)} .\left(\frac{2n}{n^{\frac{1}{n}}} -\frac{n}{n^{\frac{2}{n}}} -n\right)\\
\\
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{\log\left( 1+\left(\frac{2}{n^{\frac{1}{n}}} -\frac{1}{n^{\frac{2}{n}}} -1\right)\right)}{\left(\frac{2}{n^{\frac{1}{n}}} -\frac{1}{n^{\frac{2}{n}}} -1\right)} .\left(\frac{n^{\frac{1}{n}} -1}{n^{\frac{1}{n}}}\right)^{2} .( -n)
\end{array}$
The first term on RHS has limit equal to $\displaystyle 1\ $but the second term is giving me problem.
Please help. Thanks.
 [/INST] Note that $b_n=n^{1/n}-1\to 0$ and by Taylor series we have $$nb_n= \log n+\frac{(\log n) ^2}{2n}+ o\left(\frac{(\log n) ^2}{n}\right)=\log n+o(1)$$ and hence we have $$\log a_n=n\log (1+ 2b_n) -2\log n$$ Next we have via Taylor series $$\log a_n=2nb_n-2\log n-2nb_n^2+o(nb_n^2)=o(1)$$ (as $nb_n-\log n=o(1)$ and $nb_n^2=o(1)$) and hence $a_n\to 1$.

Alternatively let $a_n=c_n^n$ so that $$c_n=\frac{2n^{1/n}-1}{n^{2/n}}$$ and $$n(c_n-1)=-n\left(1-n^{-1/n}\right)^2=-\left(\frac{1-\exp(-(\log n) /n) }{-(\log n) /n}\right)^2\cdot\frac{(\log n) ^2}{n}$$ so that $n(c_n-1)\to 0$ and $a_n=c_n^n\to 1$ via lemma of Thomas Andrews.
</s>