<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] Here is yet another estimation of the $\lim_n\frac{(2n^{\frac 1n}-1)^n}{n^2}$.
Since
$$a_n= \frac{(2n^{\frac 1n}-1)^n}{n^2}=\Big(\frac{2n^{1/n}-1}{n^{2/n}} \Big)^n=\Big(\frac{-(1-2n^{1/n}+n^{2/n})+n^{2/n}}{n^{2/n}}\Big)^n=\Big(1-\Big(\frac{1-n^{1/n}}{n^{1/n}}\Big)^2\Big)^n$$
we obtain that
$$1\leq a_n\leq \exp\left(n\Big(\frac{1-n^{1/n}}{n^{1/n}}\Big)^2\right)$$
(Here we use the inequality $1+v\leq e^v$ for all $v$). The expression in the exponent can be rewritten as
$$n\Big(\frac{1-n^{1/n}}{n^{1/n}}\Big)^2=\frac{1}{n^{2/n}}\Big(\frac{n^{1/n}-1}{n^{-1/2}}\Big)^2$$
Since $n^{2/n}\xrightarrow{n\rightarrow0}1$, it is enough to check that $\frac{n^{1/n}-1}{n^{-1/2}}$ converges and find the limit. Here we may try L'Hospital rule, for we have a indeterminacy of the type $\frac{0}{0}$ as $n\rightarrow\infty$. Set $f(x)=x^{1/x}-1$ and $g(x)=x^{-1/2}$.
$$\frac{f(x)}{g(x)}=\frac{x^{1/x}-1}{x^{-1/2}}\sim \frac{f'(x)}{g'(x)}=2x^{1/x}\frac{\log x -1}{x^{1/2}}$$
Now $x^{1/x}\xrightarrow{x\rightarrow\infty}1$, while $\frac{\log x-1}{x^{1/2}}\xrightarrow{x\rightarrow\infty}0$ (this follows from a the well known limit $\lim_{u\rightarrow\infty}\frac{u^\alpha}{(1+p)^u}=0$ for any $p>0$ and $\alpha\in\mathbb{R}$) Consequently
(by L'Hospital theorem) $\lim_{x\rightarrow\infty}\frac{f(x)}{g(x)}=L$ exists and equals $L=0$. Putting things together, we obtain that
$$
\exp\left(n\Big(\frac{1-n^{1/n}}{n^{1/n}}\Big)^2\right)\xrightarrow{n\rightarrow\infty}1$$
Therefore, $\lim_{n\rightarrow\infty}a_n=1$.
</s>