How find this limits $\lim_{n\to\infty}\left(\sin{\frac{\ln{2}}{2}}+\sin{\frac{\ln{3}}{3}}+\cdots+\sin{\frac{\ln{n}}{n}}\right)^{1/n}$ 
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 
 A: For $k>1$ we have $0<\frac{\ln k}{k}<\frac{\pi}{2}$, so $0<\sin\frac{\ln k}{k}<\frac{\ln k}{k}$ and 
$$
\sin\frac{\ln 2}{2}<\sum_{k=1}^n\sin\frac{\ln k}{k}<\sum_{k=1}^{n}\frac{\ln k}{k}
$$
$$
\ln\sin\frac{\ln 2}{2}<\ln\left(\sum_{k=1}^n\sin\frac{\ln k}{k}\right)<\ln\left(\sum_{k=1}^{n}\frac{\ln k}{k}\right)
$$
$$
\frac{1}{n}\ln\sin\frac{\ln 2}{2}<\frac{1}{n}\ln\left(\sum_{k=1}^n\sin\frac{\ln k}{k}\right)<\frac{1}{n}\ln\left(\sum_{k=1}^{n}\frac{\ln k}{k}\right)\tag{1}
$$
Clearly
$$
\lim_{n\to\infty}\frac{1}{n}\ln\sin\frac{\ln 2}{2}=0\tag{2}
$$
On the other hand $\sum_{k=1}^{n}\frac{\ln k}{k}\sim\int_2^n \frac{\ln x}{x}\sim\ln\ln n$, so
$$
\lim_{n\to\infty}\frac{1}{n}\ln\left(\sum_{k=1}^{n}\frac{\ln k}{k}\right)
=\lim_{n\to\infty}\frac{1}{n}\ln\ln\ln n=0\tag{3}
$$
From $(1)$, $(2)$ and $(3)$ and the hamburger lemma it follows that 
$$
\frac{1}{n}\ln\left(\sum_{k=1}^n\sin\frac{\ln k}{k}\right)=0
$$
The rest is clear.
A: Since for small $x$ we have $\sin x = x+O(x^3)$, by partial summation:
$$\sum_{k=1}^{n}\sin\frac{\log k}{k}=O(1)+\sum_{k=1}^{n}\frac{\log k}{k}=\frac{1}{2}\log^2 n+O(1),$$
hence the value of the limit is just $1$.
A: 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
$$\lim_{n\to +\infty}\left(\sin{\frac{\ln{2}}{2}}+\sin{\frac{\ln{3}}{3}}+\cdots+\sin{\frac{\ln{n}}{n}}\right)^{1/n}=1.$$
