I would like to show that : $$(-1)^{n}\sqrt[n]{n}\sin(\frac{1}{n})=\dfrac{(-1)^{n}}{n}+\mathcal{O}\left(\dfrac{\ln(n)}{n^{2}} \right)$$ by starting from the left side and get the right side :
My Proof:
Note that : $$\sin(x)=x+\mathcal{O}(x^{3})$$ $$e^{x}=1+\mathcal{O}(x)$$
\begin{align*} (-1)^{n}\sqrt[n]{n}\sin\left(\frac{1}{n}\right)&= (-1)^{n}e^{\frac{1}{n}\ln\left(n\right)}\sin\left(\frac{1}{n}\right)\\ &= (-1)^{n}\left(1+\mathcal{O}\left(\dfrac{\ln(n)}{n} \right) \right)\left(\dfrac{1}{n}+\mathcal{O}\left( \dfrac{1}{n^{3}}\right) \right)\\ &= (-1)^{n}\left[\dfrac{1}{n}+\mathcal{O}\left( \dfrac{1}{n^{3}}\right)+\mathcal{O}\left(\dfrac{\ln(n)}{n^{2}}\right)+\mathcal{O}\left(\dfrac{\ln(n)}{n^{4}} \right) \right]\\ &=(-1)^{n}\left[\dfrac{1}{n}+\mathcal{O}\left(\dfrac{\ln(n)}{n^{2}}\right) \right]\\ &=\dfrac{(-1)^{n}}{n}+\mathcal{O}\left(\dfrac{\ln(n)}{n^{2}} \right)\\ \end{align*}
since : $\mathcal{O}\left(\dfrac{1}{n^{3}} \right)+\mathcal{O}\left( \dfrac{\ln(n)}{n^{4}}\right)=\mathcal{O}\left(\dfrac{\ln(n)}{n^{2}} \right)$
$$\frac{\dfrac{1}{n^3}}{\dfrac{\ln(n)}{n^2}} \xrightarrow[n\to\infty]{} 0 \implies \dfrac{1}{n^{3}}=\mathcal{O}\left( \dfrac{\ln(n)}{n^{2}} \right) $$
$$\frac{\dfrac{\ln(n)}{n^4}}{\dfrac{\ln(n)}{n^{2}}} \xrightarrow[n\to\infty]{} 0 \implies \dfrac{\ln(n)}{n^{4}}=\mathcal{O}\left( \dfrac{\ln(n)}{n^{2}} \right)$$
Is my proof correct ?