How to calculate the limit of sequence

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

$$\lim_{n\to\infty }\frac{\left(2\sqrt[n]{n}-\sqrt[n]{2}\right)^n}{n^2}=\frac12$$

Tags:

Limits

Calculus

Sequences And Series

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