Solve Sum of Infinite Series: cos(n*pi)/5^n

[APolaris]

Question says: \sum(cos(n*pi)/5^n) from 0 to infinity.

Proved that it converges: ratio test goes to abs(cos(pi*(n+1))/5cos(pi*n)) with some basic algebra. As n goes to infinity, this approaches -1/5 (absolute value giving 1/5) since cos(pi*(n+1))/cos(pi*n) is always -1, excepting the asymptotes.

Question wants to find sum. Wolfram claims sum is 5/6 and won't elaborate. How?

 

[Same-same]

Edit: Nevermind, misread what you wrote. My bad.

 

[APolaris]

The ratio test, I believe, is to use the limit as n goes to infinity of a(n+1) / a(n). So for detail:

cos (pi*(n+1))/5^(n+1) * 5^(n)/cos(pi*n).

I believe 5^n reduces with 5^(n+1) in the denominator, leaving 5 in the denominator, does it not?

 

[Dick]

The ratio test, I believe, is to use the limit as n goes to infinity of a(n+1) / a(n). So for detail:

cos (pi*(n+1))/5^(n+1) * 5^(n)/cos(pi*n).

I believe 5^n reduces with 5^(n+1) in the denominator, leaving 5 in the denominator, does it not?

Write out the first few terms of your series. You have a geometric series in disguise. That's how WA is summing it.

 

[APolaris]

Thank you.