How to calculate the limit of sequence
The question is calculate limit of sequence
$$ \lim_{n \to \infty} \frac{\left(2\,\sqrt[\Large n]{\, n\,}\, - \,\sqrt[\Large n]{\, 2\,}\right)^{n}}{n^2} $$I'm trying to simplify the equation, like divide $\,\sqrt[\Large n]{\, n\,}\,$, but can't get more. I drew the continuous function plot, which shows value tends to $0.4$.
Do any method show more details about this question ？.
$\endgroup$2 Answers
$\begingroup$Note that we have
$$\begin{align} \left(2\sqrt[n]{n}-\sqrt[n]{2}\right)^n&=\left(2e^{\frac1n \log(n)}-e^{\frac1n\log(2)}\right)^n\\\\ &=\left(1+\frac1n\log(n^2/2)+O\left(\frac{\log^2(n)}{n^2}\right)\right)^n\\\\ &=e^{n\log\left(1+\frac1n\log(n^2/2)+O\left(\frac{\log^2(n)}{n^2}\right)\right)}\\\\ &=e^{n\left(\frac1n\log(n^2/2)+O\left(\frac{\log^2(n)}{n^2}\right)\right)}\\\\ &=\frac{n^2}{2}+O\left(n\log^2(n)\right)\tag1 \end{align}.$$
Upon dividing $(1)$ by $n^2$ and letting $n\to \infty,$ we find
$$\lim_{n\to\infty }\frac{\left(2\sqrt[n]{n}-\sqrt[n]{2}\right)^n}{n^2}=\frac12$$
$\endgroup$ 13 $\begingroup$Another way to calculate your limit.
Since $\;\lim_\limits{n\to\infty}\sqrt[n]{n}=\lim_\limits{n\to\infty}\sqrt[n]{2}=1,\;$ we get that $$a_n=\frac{2\sqrt[n]{n}-\sqrt[n]{2}-\left(\sqrt[n]{n}\right)^2}{\left(\sqrt[n]{n}\right)^2}\xrightarrow{\text{as}\;n\to\infty}0$$ and $$\lim_\limits{n\to\infty}\left(1+a_n\right)^\frac{1}{a_n}=e.$$
Moreover, it results that$$\frac{(2\sqrt[n]{n}-\sqrt[n]{2})^n}{n^2}=\left[\left(1+a_n\right)^\frac{1}{a_n}\right]^{na_n}.$$
Now we are going to calculate $\;\lim_\limits{n\to\infty}(na_n)\;$.
$$na_n=-\left[\frac{n\left(\sqrt[n]{2}-1\right)+n\left(\sqrt[n]{n}-1\right)^2}{\left(\sqrt[n]{n}\right)^2}\right].$$
But $$\;\;\lim_\limits{n\to\infty}n\left(\sqrt[n]{2}-1\right)=\lim_\limits{n\to\infty}\frac{2^{\frac{1}{n}}-1}{\frac{1}{n}}=\ln2\;\;$$ and $$\lim_\limits{n\to\infty} n\left(\sqrt[n]{n}-1\right)^2=\lim_\limits{n\to\infty}\left[\left(\frac{e^{\frac{\ln n}{n}}-1}{\frac{\ln n}{n}}\right)^2\cdot\frac{\ln^2n}{n}\right]=1^2\cdot0=0$$ therefore $$\lim_\limits{n\to\infty}(na_n)=-\ln2$$ so $$\lim_\limits{n\to\infty} \frac{(2\sqrt[n]{n}-\sqrt[n]{2})^n}{n^2}=\lim_\limits{n\to\infty}\left[\left(1+a_n\right)^\frac{1}{a_n}\right]^{na_n}=e^{-\ln2}=\frac{1}{2}.$$
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