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

Question

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

Answer 1

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.

Answer 2

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

$$\lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^n\frac{\sin(i)}{i}=0$$