Question

how can i prove that the series of sin(n) is divergent?

Answer 1

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

Answer 2

any help please? tell me what are you thinking of

Answer 3

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.

Answer 4

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

Answer 5

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