Series to represent alternate between 1 and -1

[foo9008]

Homework Statement

we know that a2 , a4 ,a6 (even number ) = 0 , but when a1 , a3 , a5 (odd numbers) , the answer of an alternate between positive and negative ... in the second circle , the author represent it with (-1)^(n+1) , i don't think this is correct , this is because when n=3 , an = -2/ 3pi when n=3 , [ (-1) ^(3+1 ) ] = positive 1 , not negative 1 ! can someone explain on this ?

Homework Equations

The Attempt at a Solution

 

[stevendaryl]

I think that the author redefined n. To make it clear, you start off assuming that

f(x) = a_0 + \sum_n a_n cos(nx)

Then, for this particular problem, you find that for n > 0, then a_n = 0 unless n is odd. If n is odd, then that means that n can be written as:

n = 2n'-1

where n' = 1, 2, 3, ...

So the term -\frac{2}{\pi} \frac{cos(3x)}{1} corresponds to n=3, but it corresponds to n' = 2. In terms of n', the general term is

\frac{2}{\pi} \frac{cos((2n'-1)x)}{2n'-1} (-1)^{n'+1}