Does the series $\sum_{n=1}^{\infty}\frac{\sin n}{n}(-1)^{n}$ converges or not?

Question

From other questions we could know the series $\sum_{n=1}^{\infty}\frac{\sin n}{n}$ converges (using the Euler equaiton) and $\sum_{n=1}^{\infty}\frac{|\sin n|}{n}$ diverges (using the Euler–Maclaurin formula), but how to prove the series $\sum_{n=1}^{\infty}\frac{\sin n}{n}(-1)^{n}$ converges or not? I can not find some useful techniques. I hope some help, I am very grateful!!!

Answer 1

We are goibg to use Dirichlet's criteria for series. Let: $$a_n=\frac{1}{n}$$ This is a decreasing sequence and $\lim_{n\to +\infty}a_n=0$. Now, let: $$b_n=(-1)^n\cdot\sin(n)$$ We have to show that the sum: $$S_M=\sum_{i=0}^{M}(-1)^i\cdot \sin(i)$$ is bounded. Or, there is a $K$ indipendent from $M$, such that: $$\left|\sum_{i=0}^{M}(-1)^i\cdot \sin(i)\right|\leq K$$

Multiply the sum by $2\sin(1)$: $$2\sin(1)\cdot S_M=2\sin(1)\cdot\sum_{i=0}^{M}(-1)^i\cdot \sin(i)$$

Using the indentity $2\sin(a)\sin(b)=\cos(a-b)-\cos(a+b)$, we have:

$$2\sin(1)\cdot S_M=\sum_{i=0}^{M}(-1)^i\cdot\left(\cos(1-i)-\cos(1+i)\right)$$

Expanding some terms: $$i=0\implies \cos(1)-\cos(1)$$ $$i=1\implies -\cos(0)+\cos(2)$$ $$i=2\implies \cos(1)-\cos(3)$$ $$i=3\implies -\cos(2)+\cos(4)$$

We have a telescopic sum. In particular, it converges to: $$2\sin(1)\cdot S_M=\cos(1)-\cos(0)+\cos(M+1)-\cos(M-1)\implies \left|S_M\right|\leq \frac{2}{\sin(1)}=K$$

Thus, the series converges by Dirichlet criteria.