Show that the series cos(n) from n=1 to infinity is divergent.

Answer 1

The sum of the series ∑cos(nx) is, according to Mathworld,
N
∑cos(nx) = cos(Nx/2)sin((N+1)x/2) / sin(x/2)
n=0
(Note that the summation starts from 0, make adjustments accordingly).
The given series has x=1, or
S(1)=∑cos(n)

We see that the sum to N oscillates as N increases. Since we cannot find a value of N for whcih |T(n+1)|/|T(n)|<1 ∀n>N , we conclude that the series does not converge.

I make a difference between divergence where the sum approaches ±∞ and where the sum oscillates. I call the latter non-convergent.

Answer 2

ngf