## anonymous 5 years ago how can i prove that the series of sin(n) is divergent?

1. anonymous

$\sum_{n=0}^{\infty}\sin(n)$

2. anonymous

any help please? tell me what are you thinking of

3. anonymous

You can write the taylor series for sin(n) and use the ratio test. $\sum_{n=0}^{\infty}(-1)^n*x ^{2n+1}/(2n+1)!$ But you want to change the series index to n = 1 to perform the ratio test. You should get an infinite limit, which shows that the series diverges.

4. anonymous

if i use the ratio test, it will be convergent for all values of x actually i didn't understand your point

5. anonymous

Yeah, I see we want to sum all the sin(n)'s, not the sum of the terms of the representation of sin(n)...