Denote f ( α α ) := D KL ( ( 1 − − α α ) Q + α α P ∥ ∥ Q ) and note that D KL ( P ∥ ∥ Q ) = f ( 1 ) . The first derivative of f may be derived and evaluated as follows f ′ ( α α ) = ∑ ∑ x ∈ ∈ X ( P ( x ) − − Q ( x ) ) ( log ⁡ ⁡ ( ( 1 − − α α ) Q ( x ) + α α P ( x ) Q ( x ) ) + 1 ) = ∑ ∑ x ∈ ∈ X ( P ( x ) − − Q ( x ) ) log ⁡ ⁡ ( ( 1 − − α α ) Q ( x ) + α α P ( x ) Q ( x ) ) f ′ ( 0 ) = 0 Further derivatives may be derived and evaluated as follows f ″ ( α α ) = ∑ ∑ x ∈ ∈ X ( P ( x ) − − Q ( x ) ) 2 ( 1 − − α α ) Q ( x ) + α α P ( x ) f ″ ( 0 ) = ∑ ∑ x ∈ ∈ X ( P ( x ) − − Q ( x ) ) 2 Q ( x ) f ( n ) ( α α ) = ( − − 1 ) n ( n − − 2 ) ! ∑ ∑ x ∈ ∈ X ( P ( x ) − − Q ( x ) ) n ( ( 1 − − α α ) Q ( x ) + α α P ( x ) ) n − − 1 f ( n ) ( 0 ) = ( − − 1 ) n ( n − − 2 ) ! ∑ ∑ x ∈ ∈ X ( P ( x ) − − Q ( x ) ) n Q ( x ) n − − 1 Hence solving for D KL ( P ∥ ∥ Q ) via the Taylor expansion of f about 0 evaluated at α α = 1 yields D KL ( P ∥ ∥ Q ) = ∑ ∑ n = 0 ∞ ∞ f ( n ) ( 0 ) n ! = ∑ ∑ n = 2 ∞ ∞ 1 n ( n − − 1 ) ∑ ∑ x ∈ ∈ X ( Q ( x ) − − P ( x ) ) n Q ( x ) n − − 1 P ≤ ≤ 2 Q a.s. is a sufficient condition for convergence of the series by the following absolute convergence argument ∑ ∑ n = 2 ∞ ∞ | 1 n ( n − − 1 ) ∑ ∑ x ∈ ∈ X ( Q ( x ) − − P ( x ) ) n Q ( x ) n − − 1 | = ∑ ∑ n = 2 ∞ ∞ 1 n ( n − − 1 ) ∑ ∑ x ∈ ∈ X | Q ( x ) − − P ( x ) | | 1 − − P ( x ) Q ( x ) | n − − 1 ≤ ≤ ∑ ∑ n = 2 ∞ ∞ 1 n ( n − − 1 ) ∑ ∑ x ∈ ∈ X | Q ( x ) − − P ( x ) | ≤ ≤ ∑ ∑ n = 2 ∞ ∞ 1 n ( n − − 1 ) = 1 P ≤ ≤ 2 Q a.s. is also a necessary condition for convergence of the series by the following proof by contradiction. Assume that P > 2 Q with measure strictly greater than 0 . It then follows that there must exist some values ϵ ϵ > 0 , ρ ρ > 0 , and U < ∞ ∞ such that P ≥ ≥ 2 Q + ϵ ϵ and Q ≤ ≤ U with measure ρ ρ . The previous proof of sufficiency demonstrated that the measure 1 − − ρ ρ component of the series where P ≤ ≤ 2 Q is bounded, so we need only concern ourselves with the behavior of the measure ρ ρ component of the series where P ≥ ≥ 2 Q + ϵ ϵ . The absolute value of the n th term of this component of the series is then lower bounded by 1 n ( n − − 1 ) ρ ρ ( 1 + ϵ ϵ U ) n , which is unbounded as n → → ∞ ∞ , so the series diverges.