The sum from n=1 to infinity of cos(npi/3)/n!

Does this absolutely converge, conditionally converge, or diverge?

Group every 6 terms (1 cycle) together and consider as an aggregate term for a new series, ΣQn. Assume that all the terms are positive because of the n! in the denominator.

If you can prove that
Qn+1/Qn < 1, then
by the ratio test, the (new) series is absolutely convergent.

Can you take it from here?

To determine whether the given series absolutely converges, conditionally converges, or diverges, we can apply the ratio test.

The ratio test states that if the limit of the absolute value of the ratio between successive terms of a series is less than 1, then the series absolutely converges. If the limit is greater than 1 or it does not exist, the series diverges. If the limit is equal to 1, the ratio test is inconclusive.

Let's calculate the limit:

lim┬(n→∞)⁡|(cos((n+1)π/3)/(n+1)!) / (cos(nπ/3)/n!)|

Simplifying the expression:

lim┬(n→∞)⁡|(cos((n+1)π/3) * n!) / (cos(nπ/3) * (n+1)!)|

cos((n+1)π/3) can be written as cos(nπ/3 + π/3) using the periodicity of cosine function.

Let's use the trigonometric identity: cos(A + B) = cos(A) * cos(B) - sin(A) * sin(B):

cos(nπ/3 + π/3) = cos(nπ/3) * cos(π/3) - sin(nπ/3) * sin(π/3)

Simplifying further:

= (cos(nπ/3) * 1/2) - (sin(nπ/3) * √3/2)

Now, substituting the expression into our limit:

lim┬(n→∞)⁡|(cos(nπ/3) * n! * 2) / (cos(nπ/3) * (n+1)!) * 1/(√3/2) * 2/1|

Simplifying:

lim┬(n→∞)⁡|(2 * n!) / ((n+1)! * √3)|

Using the factorial property: (n+1)! = (n+1) * n!

lim┬(n→∞)⁡|(2 * n!) / ((n+1) * n! * √3)|

The n! terms cancel out:

lim┬(n→∞)⁡|2 / (√3 * (n+1))|

As n approaches infinity, the denominator (√3 * (n+1)) increases without bound, while the numerator (2) remains constant.

Therefore, the limit is 0:

lim┬(n→∞)⁡|2 / (√3 * (n+1))| = 0

Since the limit is less than 1, according to the ratio test, the given series absolutely converges.