Show that the series cos(n) from n=1 to infinity is divergent.
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.
ngf
To show that the series ∑cos(n) from n=1 to infinity is divergent, we can use the divergence test. The divergence test states that if the terms of a series do not approach zero, then the series diverges.
In this case, let's consider the sequence of terms {cos(n)}. If the series converges, then the terms of the series must approach zero.
However, since the cosine function oscillates between -1 and 1, the terms of the sequence {cos(n)} do not approach zero. Instead, they alternate between these two values as n increases.
Therefore, by the divergence test, the series ∑cos(n) from n=1 to infinity is divergent.
To determine the convergence or divergence of the series sum(cos(n)) from n=1 to infinity, we can use the divergence test.
The divergence test states that if the terms of a series do not tend to zero as n approaches infinity, then the series diverges.
In this case, let's consider the terms of the series: cos(n).
To show that the series diverges, we need to show that the terms of the series do not approach zero as n approaches infinity.
We know that the cosine function oscillates between -1 and 1, and it never approaches a single value as n increases. Specifically, cos(n) oscillates between -1 and 1 for different values of n.
Therefore, the terms of the series cos(n) do not tend to zero as n approaches infinity.
Since the terms of the series do not approach zero, we can conclude that the series sum(cos(n)) from n=1 to infinity is divergent.