How find this limits $\lim_{n\to\infty}\left(\sin{\frac{\ln{2}}{2}}+\sin{\frac{\ln{3}}{3}}+\cdots+\sin{\frac{\ln{n}}{n}}\right)^{1/n}$

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.

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 $1$.

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 $$\lim_{n\to +\infty}\left(\sin{\frac{\ln{2}}{2}}+\sin{\frac{\ln{3}}{3}}+\cdots+\sin{\frac{\ln{n}}{n}}\right)^{1/n}=1.$$