Finding $\lim \frac{(2n^{\frac 1n}-1)^n}{n^2}$.

Question

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.

Answer 1

It's best to write it as :- $$\lim_{n\to\infty}\left(\frac{2n^{\frac{1}{n}}-1}{n^{\frac{2}{n}}}\right)^{n}$$.

Now this is $\displaystyle \lim_{n\to\infty}\left(2n^{\frac{-1}{n}}-n^{\frac{-2}{n}}\right)^{n}$ which is $1^{\infty}$ form.

Now if $f(x)^{g(x)}$ has $1^{\infty}$ form then you can evaluate the following limit and they will be equal:-

$e^{\lim_{x\to a}(f(x)-1)g(x)}$ and it works for the infinite limit as well.

So this becomes:-

$$\text{exp}\left(\lim_{n\to\infty}\frac{\left(2n^{\frac{-1}{n}}-n^{\frac{-2}{n}}-1\right)}{\frac{1}{n}}\right)$$

Now substitute $h=\frac{1}{n}$ and evaluate the limit at $h\to 0^{+}$ . Note :- I will just use $h\to 0$ to denote $h\to 0^{+}$ as it is cumbersome to write that everytime.

$$\text{exp}\left(\lim_{h\to 0}\frac{\left(2h^{h}-h^{2h}-1\right)}{h}\right)$$

Now you can use L'hospital as we know $h^{h}$ tends to $1$ as $h\to 0^{+}$.

So we get :-

$$\text{exp}\left(\lim_{h\to 0}\,\, 2h^{h}(1-\ln(h)) - (2-2\ln(h))h^{2h}\right)$$

$$=\text{exp}\left(\lim_{h\to 0}\,\, 2h^{h}-2+2(h^{2h}-h^{h})\ln(h)\right)$$ .

Now we just need to show that $$\displaystyle\lim_{h\to 0}2(h^{2h}-h^{h})\ln(h)=\lim_{h\to0}2h^{h}(h^{h}-1)\ln(h)=0$$

Or in other words, we need to prove $\displaystyle\lim_{h\to0}(h^{h}-1)\ln(h)=0$ , as $2h^{h}$ would tend to $2$.

Notice that for we can write $h^{h}=e^{h(\ln(h))}$.

Now we see that the limit is just :-

$$\lim_{h\to 0}(h^{h}-1)\ln(h)=\lim_{h\to 0}(e^{h(\ln(h))}-1)\ln(h)$$

$$=\lim_{h\to 0} \left(h\ln(h)+\frac{h^{2}(\ln(h))^{2}}{2!}+o(h^{3}(ln^{3}(h))\right)\ln(h)$$

But we know $\lim_{h\to 0} h^{m}\ln^{m+1}(h)=0$ for $\,m> 0$ (Prove this. Youtube has also many videos showing this. It's nothing too special and a well known result.

hence finally you have your answer as $\displaystyle \text{exp}(0)=1$

Answer 2

Power Series Approach

Using the power series $e^x=1+x+O\!\left(x^2\right)$, we get $$ \begin{align} 2n^{1/n}-1 &=2e^{\frac1n\log(n)}-1\tag{1a}\\[6pt] &=2\left(1+\frac1n\log(n)+O\!\left(\frac{\log(n)^2}{n^2}\right)\right)-1\tag{1b}\\ &=1+\frac2n\log(n)+O\!\left(\frac{\log(n)^2}{n^2}\right)\tag{1c} \end{align} $$ and $$ \begin{align} n^{2/n} &=e^{\frac2n\log(n)}\tag{2a}\\[6pt] &=1+\frac2n\log(n)+O\!\left(\frac{\log(n)^2}{n^2}\right)\tag{2b} \end{align} $$ Therefore, $$ \begin{align} \frac{\left(2n^{1/n}-1\right)^n}{n^2} &=\left(\frac{2n^{1/n}-1}{n^{2/n}}\right)^n\tag{3a}\\ &=\left(1+O\!\left(\frac{\log(n)^2}{n^2}\right)\right)^n\tag{3b}\\ &=1+O\!\left(\frac{\log(n)^2}n\right)\tag{3c} \end{align} $$ Explanation:
$\text{(3a)}$: $n^2=\left(n^{2/n}\right)^n$
$\text{(3b)}$: apply $(1)$, $(2)$, and $\frac{1+x+O\left(x^2\right)}{1+x+O\left(x^2\right)}=1+O\!\left(x^2\right)$
$\text{(3c)}$: Binomial Theorem

Thus, $$ \lim_{n\to\infty}\frac{\left(2n^{1/n}-1\right)^n}{n^2}=1\tag4 $$

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More Elementary Approach

For $n\ge12$, $$ 4(n-1)(n-2)-3n^2=n(n-12)+8\gt0\tag5 $$ Thus, $\frac43(n-1)(n-2)\gt n^2$. Then the Binomial Theorem says, $$ \left(1+2n^{-2/3}\right)^n \ge1+2n\cdot n^{-2/3}+2n(n-1)\cdot n^{-4/3}+\overbrace{\frac43n(n-1)(n-2)\cdot n^{-2}}^{\ge n}\tag6 $$ Therefore, $$ \begin{align} 1-n^{-1/n} &\le n^{1/n}-1\tag{7a}\\[6pt] &\le2n^{-2/3}\tag{7b} \end{align} $$ Explanation:
$\text{(7a)}$: $1\le n^{1/n}$
$\text{(7b)}$: $(6)$ says that $\left(1+2n^{-2/3}\right)^n\ge n$

Finally, $$ \begin{align} \frac{\left(2n^{1/n}-1\right)^n}{n^2} &=\left(\frac{2n^{1/n}-1}{n^{2/n}}\right)^n\tag{8a}\\ &=\left(1-\frac{\left(n^{1/n}-1\right)^2}{n^{2/n}}\right)^n\tag{8b}\\[3pt] &=\left(1-\left(1-n^{-1/n}\right)^2\right)^n\tag{8c}\\[9pt] &\ge1-n\left(1-n^{-1/n}\right)^2\tag{8d}\\[12pt] &\ge1-4n^{-1/3}\tag{8e} \end{align} $$ Explanation:
$\text{(8a)}$: $n^2=\left(n^{2/n}\right)^n$
$\text{(8b)}$: expand the square
$\text{(8c)}$: $\frac{n^{1/n}-1}{n^{1/n}}=1-n^{-1/n}$
$\text{(8d)}$: Bernoulli's Inequality, applicable since $\left(1-n^{-1/n}\right)^2\in[0,1]$
$\text{(8e)}$: apply $(7)$

$\text{(8c)}$, for the upper bound, and $\text{(8e)}$, for the lower bound, yield $$ 1-4n^{-1/3}\le\frac{\left(2n^{1/n}-1\right)^n}{n^2}\le1\tag9 $$ to which we can apply the Squeeze Theorem to get $(4)$.

Answer 3

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 $a_n\to 1$.

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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.

Answer 4

Use these two results:

We can use that for $|\alpha x|>>1$, $$(1+\alpha)^x \sim e^{\alpha x} \tag{1}$$

We have the basic limit for logarithm as; $$\lim_{n \rightarrow \infty}n\left(a^{1/n}-1\right)=\log a \hspace{5px}\text{, for all positive } a$$ for $a =n$ itself we have: $$\lim_{n \rightarrow \infty}n\left(n^{1/n}-1\right)=\log n \tag{2}$$

$$\begin{align}\lim_{n\rightarrow \infty}\frac{\left(2n^{1/n}-1\right)^{n}}{n^{2}}&=\lim_{n\rightarrow \infty}\frac{\left(1+2\left(n^{1/n}-1\right)\right)^{n}}{n^{2}} \\ &=\lim_{n\rightarrow \infty}\frac{\exp\left(2n\left(n^{1/n}-1\right)\right)}{n^{2}} \tag{By Eq. 1 & 2}\\ &=\lim_{n\rightarrow \infty}\frac{\exp\left(2\log n\right)}{n^{2}} \tag{By Eq. 2}\\ &=1\end{align}$$

Answer 5

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, $\lim_{n\rightarrow\infty}a_n=1$.