analysis - How find this limits $lim_{ntoinfty}left(sin{frac{ln{2}}{2}}+sin{frac{ln{3}}{3}}+cdots+sin{frac{ln{n}}{n}}right)^{1/n}$

Find this limit
$$\lim_{n\to\infty}\left(\sin{\dfrac{\ln{2}}{2}}+\sin{\dfrac{\ln{3}}{3}}+\cdots+\sin{\dfrac{\ln{n}}{n}}\right)^{1/n}$$

My idea:use
$$x=e^{\ln{x}}$$
so we only find
$$\lim_{n\to \infty}\dfrac{\ln{\left(\sin{\dfrac{\ln{2}}{2}}+\sin{\dfrac{\ln{3}}{3}}+\cdots+\sin{\dfrac{\ln{n}}{n}}\right)}}{n}$$
then
$$\lim_{n\to\infty}\dfrac{\ln{\left(\sin{\dfrac{\ln{2}}{2}}+\sin{\dfrac{\ln{3}}{3}}+\cdots+\sin{\dfrac{\ln{(n+1)}}{n+1}}\right)}-\ln{\left(\sin{\dfrac{\ln{2}}{2}}+\sin{\dfrac{\ln{3}}{3}}+\cdots+\sin{\dfrac{\ln{n}}{n}}\right)}}{(n+1)-n}=\ln{\left(\sin{\dfrac{\ln{2}}{2}}+\sin{\dfrac{\ln{3}}{3}}+\cdots+\sin{\dfrac{\ln{(n+1)}}{n+1}}\right)}-\ln{\left(\sin{\dfrac{\ln{2}}{2}}+\sin{\dfrac{\ln{3}}{3}}+\cdots+\sin{\dfrac{\ln{n}}{n}}\right)}$$
then I can't works,Thank you