We know the derivative of eˣ is eˣ.
If we write eˣ=a₀+a₁x+a₂x²+...+a_nx^n
When x=0, 1=a₀.
Now differentiate:
eˣ=a₁+2a₂x+...+na_nx^(n-1)
and put x=0:
1=a₁.
Differentiate again:
eˣ=2a₂+...n(n-1)x^(n-2).
When x=0, 1=2a₂.
If we repeatedly differentiate, and put x=0 to solve for a_n, we end up with the series:
1+x+x²/2!+x³/3!+...+x^n/n! to infinity.
Put x=1 and e=1+1+1/2!+1/3!+...+1/n!.
e-2=1/2!+1/3!+...
So the general term is 1/(n+1)!, the required series.
Therefore (b) is the answer.