If an=∑nn! then the sum of series ∑an is
e
e-1
3e2
e2
Explanation for the correct option:
Step 1. Find the value of ∑an:
Given, an=∑nn!
=n(n+1)2n!=(n+1)2(n-1)!=n-1+22(n-1)!=121(n-2)!+1(n-1)!
Step 2. Find the values of a1,a2,a3 and so on:
a1=1
a2=1210!+11!
a3=1211!+12!
∴∑an=1210!+11!+12!+....+10!+11!+12!+...=12e+e=3e2 ∵10!+11!+12!+....=e
Hence, Option ‘C’ is Correct.