TITLE: How find this limits $\lim_{n\to\infty}\left(\sin{\frac{\ln{2}}{2}}+\sin{\frac{\ln{3}}{3}}+\cdots+\sin{\frac{\ln{n}}{n}}\right)^{1/n}$
QUESTION [10 upvotes]: 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

REPLY [3 votes]: We have:
$$\sin{\frac{\ln{2}}{2}}<\sin{\frac{\ln{2}}{2}}+\sin{\frac{\ln{3}}{3}}+\cdots+\sin{\frac{\ln{n}}{n}}<n$$
$$\left(\sin{\frac{\ln{2}}{2}}\right)^{\frac{1}{n}}<\left(\sin{\frac{\ln{2}}{2}}+\sin{\frac{\ln{3}}{3}}+\cdots+\sin{\frac{\ln{n}}{n}}\right)^{\frac{1}{n}}<n^\frac{1}{n}$$
It follows that
$$\lim_{n\to +\infty}\left(\sin{\frac{\ln{2}}{2}}+\sin{\frac{\ln{3}}{3}}+\cdots+\sin{\frac{\ln{n}}{n}}\right)^{1/n}=1.$$