Evaluate:
∞ n2 - 2
Σ ---------
n=1 (n+2)!
If you used a computer, STOP! You don't need that level of overkill. This is essentially a college textbook problem.
Let's start with the series expansion of e as given: e = sum {n=0 to inf} 1/n!
Then look at sum {n=1 to inf} n/n!. After canceling out a factor of n, we have {n=1 to inf} 1/(n-1)! which is just our series for e, but with the index offset by one.
Similarly, look at sum {n=2 to inf} n*(n-1)/n!. After canceling out factors n and n-1 we have {n=2 to inf} 1/(n-2)!. Which again is our series for e but with the index offset by 2.
Now onto the question at hand: sum {n=1 to inf} (n^2-2)/(n+2)!
First I want to reindex this to match the n! in our earlier series. Then we have sum {n=3 to inf} (n^2-4n+2)/n!
Now to split this up to more closely match our knows series.
The summand can be decomposed into (n^2-4n+2)/n! = n*(n-1)/n! - 3*n/n! + 2/n!
Then plugging this into the summation and separating the parts into three separate summations yields:
sum {n=3 to inf} n*(n-1)/n! - sum {n=3 to inf} 3*n/n! + sum {n=3 to inf} 2/n!
Each of these sums are missing a few terms, so we can just add them back in if we subtract them outside of the summations. Then the series constructed at the beginning can be used to simplify:
sum {n=3 to inf} n*(n-1)/n! = -2*(2-1)/2! + sum {n=2 to inf} n*(n-1)/n! = e-1
sum {n=3 to inf} 3*n/n! = -3*1/3! - 3*2/2! + 3*sum {n=1 to inf} n/n! = 3e-6
sum {n=3 to inf} 2/n! = -2/0! - 2/1! - 2/2! + 2*sum {n=0 to inf} 1/n! = 2e-5
Now all that is left to do is add up the three sums: (e-1) - (3e-6) + (2e-5) = 0.
What if the -2 was left out of the original numerator? Then we'd end up with 2e-5~=0.43656....
Analytic Solution
