How do you solve the series from k=2 to infinity of (9/k)-(9/(k+2))

Need help solving the series from k=2 to infinity of (9/k)-(9/k+2). Would be greatly appreciated.
http://www.wolframalpha.com/input/?i=series+from+k%3D2+to+infinty+of+%289%2Fk%29-%289%2F%28k%2B2%29%29
That's the series I want. I need to see the steps to get to that answer.

The series is telescoping. Look at the Nth partial sum---the sum from 2 to N.

N
Σ 9/k - 9/(k+2) =
k=2

= 9[(1/2 - 1/4) + (1/3 - 1/5) + (1/4 - 1/6) + (1/5 - 1/7) + ... + (1/N - 1/(N+1))]

Notice that most of the terms cancel because they appear both added and subtracted. The only terms that don't cancel are 1/2 + 1/3 - 1/(N+1) - 1/(N+2). So

N
Σ 9/k - 9/(k+2) = 9(1/2 + 1/3 - 1/(N+1) - 1/(N+2)).
k=2

Take the limit as N->∞. This is the sum of the series. It is

∞
Σ 9/k - 9/(k+2) = 9(1/2 + 1/3 - 0) = 15/2
k=2