Ive done that but i cant get the answer i think my arithmetic is wrong.

hmmm...

\(\displaystyle \frac{1}{ \pi } [ \frac{1}{n+1} (-1)^n - \frac{1}{n-1} (-1)^n + \frac{1}{n+1} - \frac{1}{n-1} ] \)

Lets work with the inside for now,

\(\displaystyle \frac{1}{n+1} (-1)^n - \frac{1}{n-1} (-1)^n + \frac{1}{n+1} - \frac{1}{n-1} \)

Group terms

\(\displaystyle \frac{1}{n+1} ( 1 +(-1)^n) - \frac{1}{n-1} ( 1 + (-1)^n) \)

Note that if \(\displaystyle n = odd \) then the above must equal 0. So n must be even, which means

\(\displaystyle \frac{1}{n+1} ( 1 +(-1)^n) - \frac{1}{n-1} ( 1 + (-1)^n) \to \frac{2}{n+1} - \frac{2}{n-1} \)

This gives us

\(\displaystyle \frac{ 2(n-1) -2(n+1) }{(n+1)(n-1)} \)

\(\displaystyle - \frac{ 4} {(n+1)(n-1)} \)

So my answer will be,

\(\displaystyle - \frac{4}{ \pi} \frac{ 1} {(n+1)(n-1)} \)

Is this the same as theres? Lets change it around

If you sub in \(\displaystyle n = 2 \) into their equation and mine you arrive at \(\displaystyle - \frac{4}{3 \pi } \)

So I think they are equal. let's equate them

\(\displaystyle - \frac{4}{ \pi} \frac{ 1} {(n+1)(n-1)} = -\frac{2}{ \pi } \frac{ (-1)^n +1 } {n^2 - 1} \)

\(\displaystyle 2 \frac{ 1} {n^2 - 1} = \frac{ (-1)^n + 1 }{n^2 - 1} \)

Note how these are the same if n=even!

However, if n=odd the right side is equal to 0, which is the same result as I achieved, however i had the stipulation before making my equation that n cannot be odd.

Thus, left side = right side!