Can you show me how you got the [math]\frac{2}{2n+1}*sin(2n+1)x[/math]?
In your post #3 you have shown that only the odd index of summation matters, ie \(\displaystyle n = 1,3,5,7....\)
The trick is that if we want only even indices we change \(\displaystyle n\) to \(\displaystyle 2n\)
And
If we want only odd indices we change \(\displaystyle n\) to \(\displaystyle 2n + 1\)
Why?
Because \(\displaystyle 2n + 1\) allows us to get the odd indices again. Here are some calculations:
\(\displaystyle 2(0) + 1 = 1\)
\(\displaystyle 2(1) + 1 = 3\)
\(\displaystyle 2(2) + 1 = 5\)
\(\displaystyle 2(3) + 1 = 7\)
Aren't these the same as \(\displaystyle n = 1,3,5,7....\)?
The second part is that if we change one \(\displaystyle n\) to \(\displaystyle 2n + 1\), we have to change all other \(\displaystyle n\)'s to \(\displaystyle 2n + 1\).
In the original summation, we have:
\(\displaystyle \sum_{n=1}\frac{(-1)^{n+1} + 1}{n}\sin nx\)
It means that we have two \(\displaystyle n\)'s. So we have to change them both. And we get:
\(\displaystyle \sum_{n=0}\frac{2}{2n + 1}\sin (2n + 1)x\)
And because we care only for the odd \(\displaystyle n\), \(\displaystyle (-1)^{n+1} + 1\) is always \(\displaystyle = 2\)