# Homework Help: Find the Sum of this Alternating Series

1. Oct 29, 2014

### RJLiberator

1. The problem statement, all variables and given/known data
Find the sum of
starts at 0 to infinity ∑ (cos(k*pi))/pi^k

First, I determined that it does, indeed, converge with the alternating series test.
Second, I found the answer to be pi/(1+pi) via wolfram alpha.

But I am at a loss on how to find the answer here.

This is a geometric series, so I am trying to find the r value. If I can do that, then I can use the trick of first term/(1-r)

Correct?

2. Oct 29, 2014

### Staff: Mentor

It might be helpful to write the series without the cosine factor. All it contributes is +/- 1 each time.

3. Oct 29, 2014

### LCKurtz

Why don't you just divide the $k+1$ term by the $k$ term to get $r$?

4. Oct 29, 2014

### RJLiberator

So, with cos(k*pi) out, that would leave 1/pi^k. Therefore 1/pi^k = r? Hm.

k+1 term? Ok, let me try what I think you are saying:

k+1 term = value of 1, so k=1 and therefore cos(1pi)/pi^1 = -1/pi and the original k =0 term = cos(0pi)/pi^0 = 1. So, (-1/pi)/1 = -1/pi.

Hm, so if r = -1/pi then we can plug that into the first term/(1-r) equation, lets see:
First term = 1
(1-(-1/pi) = 1+1/pi
1/(1+1/pi)

If you multiple everything by pi, to simplify you get the answer of pi/pi+1 which is the correct answer.

Why did this work? Is there a trick I am not thinking of?

Thank you all. It seems I may have made a mechanical error with Mark44's suggestion. LcKurtz, do you know the definition/term of what you suggested so i can research it?

5. Oct 29, 2014

### Staff: Mentor

No it's not. $\frac{\pi}{\pi} + 1 = 2$.

I'm being picky, but you need parentheses like you used in your first post here. IOW, like this: π/(π + 1).

6. Oct 29, 2014

### RUber

$\cos k \pi =(-1)^k$ so you would have $\sum_k \left(\frac{-1}{\pi}\right)^k$
Mark44's suggestion should have led you to the same conclusion for r.

7. Oct 29, 2014

### LCKurtz

A geometric series looks like $$\sum_{k=0}^\infty ar^k$$where $r$ is the common ratio. The $k+1$ term is $ar^{k+1}$ and the $k$ term is $ar^k$. When you divide them you get the common ratio $r$, no matter what $k$ is. What you did is divide the $k=1$ term by the $k=0$ term. That doesn't tell you anything unless you already know it is a geometric series. What you did gave you the correct ratio because it is in fact a geometric series, but the correct way to get the ratio and show it is a geometric series is divide the $k+1$ term by the $k$ term and observe you always get the same answer.

8. Oct 30, 2014

### RJLiberator

Excellent analysis, friends. I appreciate the words. Quite frankly, I am getting addicted to this series math.