Finding $\lim{_{n \to \infty } } \frac{\frac{\sin(1)}{1}+\frac{\sin(2)}{2}+\frac{\sin(3)}{3}+...+\frac{\sin(n)}{n} }{n}$ I tried to calculate this limit:

$$\lim{_{n \to \infty  } } \dfrac{\dfrac{\sin(1)}{1}+\dfrac{\sin(2)}{2}+\dfrac{\sin(3)}{3}+...+\dfrac{\sin(n)}{n} }{n}$$

I've tried to compare to other term such as $\dfrac{1}{n^2}$ or to use Dirichlet test but have no other ideas.
Thanks
 A: We have
$$\lim{_{n \to \infty } } \frac{\frac{\sin(1)}{1}+\frac{\sin(2)}{2}+\frac{\sin(3)}{3}+...+\frac{\sin(n)}{n} }{n}=\lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^n\frac{\sin(i)}{i}$$
But
$$\left|\sum_{i=1}^n\frac{\sin(i)}{i}\right|\leq\sum_{i=1}^n\frac{|\sin(i)|}{i}\leq  \sum_{i=1}^n\frac{1}{i}=H_n$$
where $H_i$ is the $i$th Harmonic number. However, this is well known to be bounded by
$$H_n<1+\ln(n+1)$$
for large enough $n$. Thus
$$\lim_{n\to\infty}\left|\frac{1}{n}\sum_{i=1}^n\frac{\sin(i)}{i}\right|\leq \lim_{n\to\infty}\frac{1+\ln(n+1)}{n}=0$$
Since the absolute value of the expression goes to zero, we conclude
$$\lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^n\frac{\sin(i)}{i}=0$$
A: The series $\sum_{i=1}^{\infty} \frac{\sin(i)}{i}$ converge (by Abel's test) to some number $S$. Therefore, 
$$\lim_{n\to \infty} \frac{1}{n}\sum_{i=1}^{n}\frac{\sin(i)}{i} = 0$$ 
Note: 


*

*You do not need to compute $S$. 

*Another way to solve this problem is by applying Stolz lemma.

*A slightly harder version of this problem could be
$$\lim_{n\to \infty} \frac{1}{n}\sum_{i=1}^{n}\frac{|\sin(i)|}{i} $$ 
QC_QAOA has an answer for this. 

