difference method with partial fractions

[look416]

1. The problem statement, all variables and given/known data

Verify that, for all positive values of n,
1/(n+2)(2n+3) -1/((n+3)(2n+5))=(4n+9)/((n+2)(n+3)(2n+3)(2n+5))
For the series
∑_(n=0)^N▒(4n+9)/((n+2)(n+3)(2n+3)(2n+5))
Find
The sum to N terms,
The sum to infinity.


2. Relevant equations

no

3. The attempt at a solution

i had solved the first sextion which is to verification
however, when it came to the difference method solving, i had a huge problems
therefore, any solutions?

[benorin]

Google telescoping series

[look416]

most of the google doesnt reveal the method of solving series using difference method ==
i have tried it
or maybe you can give me the links if you have found it

[Fightfish]

\frac{1}{(n+2)(2n+3)} = \frac{2}{2n+3} - \frac{1}{n+2}
\frac{1}{(n+3)(2n+5)} = \frac{2}{2n+5} - \frac{1}{n+3}
I think this may be the key issue bugging you when trying to use the difference method without simplifying the expressions further.

[look416]

\frac{1}{(n+2)(2n+3)} = \frac{2}{2n+3} - \frac{1}{n+2}
\frac{1}{(n+3)(2n+5)} = \frac{2}{2n+5} - \frac{1}{n+3}
I think this may be the key issue bugging you when trying to use the difference method without simplifying the expressions further.

well i have found the answer
thanks for the help
my fault is i did not classified them correctly
as i should put 1/(n+3) and 1/(n+2) together as well as 1/(2n+3) and 2/(2n+5) together instead of i straight use the result to do difference method thats y i cant find the answer!
Anyways thanks for your help