Given:

The sequence {an} is defined recursively by the equations a0=a1=1 , n(n−1)an=(n−1)(n−2)an−1−(n−3)an−2 .

Calculation:

Rewrite the above recurrence relation as follows.

  n(n−1)an=(n−1)(n−2)an−1−(n−3)an−2an=(n−1)(n−2)an−1−(n−3)an−2n(n−1)

Let n=2

Substitute 2 for n in the above equation.

  a2=(2−1)(2−2)a2−1−(2−3)a2−22(2−1)=1(0)a2−1−(2−3)a2−22{2−1}=a02=12

Therefore, the value of a2=12! .

Let n=3

Substitute 3 for n in the above equation.

  a3=(3−1)(3−2)a3−1−(3−3)a3−23(3−1)=2(1)a2−(0)a13(2)=a23=16

Therefore, the value of a3=13! .

Let n=4

Substitute 4 for n in the above equation.

  a4=(4−1)(4−2)a4−1−(4−3)a4−24(4−1)=3(2)a3−(1)a24(3)=6a3−a212=124

Therefore, the value of a4=14! .

Repeating like this it will get an=1n! .

Now find the sum of ∑an=∑1n! .

  ∑an=∑1n!=1+12!+13!+⋯=e

Therefore, the sum of ∑n→0∞an is e .