Originally posted by Jirakon
I think it's 1.
I know it's 1, but I don't know how.
The clue didn't help then?
If we write f(x) = SUM x^(n+2)/[n!(n+2)], note that f(1) gives the series you want.
df/dx = SUM x^(n+1)/n! = x(SUM x^n/n!)
You might recognise last sum: SUM x^n/n! = e^x
so df/dx = xe^x
=> f = (x - 1)e^x + c
To calculate c, set x = 0. We know f(0) = 0, so c = 1
=> f(x) = (x - 1)e^x + 1
=> f(1) = 1
One of those cases where the easiest way to solve a problem is actually to solve a more general problem, which has a special case that matches your original problem.