2 Verify that, for all positive values of \(n\),
$$\frac { 1 } { ( n + 2 ) ( 2 n + 3 ) } - \frac { 1 } { ( n + 3 ) ( 2 n + 5 ) } = \frac { 4 n + 9 } { ( n + 2 ) ( n + 3 ) ( 2 n + 3 ) ( 2 n + 5 ) } .$$
For the series
$$\sum _ { n = 1 } ^ { N } \frac { 4 n + 9 } { ( n + 2 ) ( n + 3 ) ( 2 n + 3 ) ( 2 n + 5 ) }$$
find
- the sum to \(N\) terms,
- the sum to infinity.