Prove by Cauchy Test that the sequence $X_n= \frac{\sin(1)}{2}+\frac{\sin(2)}{2^2}+...+\frac{\sin(n)}{2^n}$ converges. Prove that the sequence $X_n= \frac{\sin(1)}{2}+\frac{\sin(2)}{2^2}+...+\frac{\sin(n)}{2^n}$ converges.
I will add a photo of my solution, and I would be glad if someone gives me a hint to continue from where i got stuck. I want to use the sum of infinite series at the end of my solution, which means that $S_{\infty}= \frac{a_1}{1-q}$. put I've noticed that I have only $P$ terms in the sum, which means my $|q|^n$ doesn't cnverge to zero. What am I missing here? 
Note: you can ignore the wrong move I did in line 4, by removing the absolute value. but it still true that $|sin(n)|\le 1$.

 A: You can avoid the Cauchy test by using the comparison test:
$$
\sum_{n=1}^{\infty} \left| \frac{\sin(n)}{2^n} \right|
\le
\sum_{n=1}^{\infty} \frac{1}{2^n}
= 1
$$
To continue your work on the Cauchy test:
$$
\frac1{2^{n+1}} + \cdots + \frac1{2^{n+p}}
=
\frac1{2^{n}} (1-\frac1{2^{p}})
\le
\frac1{2^{n}} \to 0
$$
A: $\sum_{r=0}^n\dfrac{\sin n}{2^n}$
=imaginary of $\sum_{r=0}^n\dfrac{e^{in}}{2^n}$
Now $$\sum_{r=0}^n\dfrac{e^{in}}{2^n}=\sum_{r=0}^n\left(\dfrac{e^i}2\right)^n=\dfrac1{1-\dfrac{e^i}2}=\dfrac1{1-\dfrac{\cos1+i\sin 1}2}$$
Can you separate out the imaginary part to take it home from here?
A: You can continue in this way to go for the proof for Cauchy sequence.
This is what you have to show:
$$\left|X_m-X_n\right|< \epsilon$$
Say $m=n+1$
$$\left|X_{n+1}-X_n\right|=\left|\sum_{r=1}^{n+1}\frac{\sin (r)}{2^r}-\sum_{r=1}^{n}\frac{\sin (r)}{2^r}\right|=\left|\frac{\sin (n+1)}{2^{n+1}}\right|$$
$$<\left|\frac{1}{2^{n+1}}\right|=\left(\frac{1}{2}\right)^{n+1}<\epsilon \,\ \forall \,\ n \in \mathbb{N}$$
