Weird limits

Calculus Level 5

lim ⁡ n → ∞ ( n 7 ) p 7 q n − 7 = e A B C D ! \large \lim _{ n\rightarrow \infty }{ \binom{n}{7} { p }^{ 7 }{ q }^{ n-7 } }=\frac {{e}^{A}{B}^{C}}{D!}\

Given the above, where p + q = 1 p+q=1 and n p = 5 np=5 , find A + B + C + D A+B+C+D .

Bonus: Generalize for r r and m m in place of 7 and 5 respectively.

Notation: ( n r ) = n ! ( n − r ) ! r ! \displaystyle \binom{n}{r}=\frac{n!}{(n-r)!r!} denotes the binomial coefficient .


The answer is 14.

Rishi Sharma by Rishi Sharma , 32, India

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1 solution

The equation expresses the relationship that as n → ∞ n \to \infty and p → 0 p \to 0 the binomial distribution (the LHS) converges to the Poisson distribution (the RHS).

lim ⁡ n → ∞ ( n r ) p r ( 1 − p ) n − r = e − λ λ r r ! = e − m m r r ! where the mean is λ = m = n p . \begin{aligned} \lim_{n \to \infty} \binom nr p^r(1-p)^{n-r} & = \frac {e^{-\lambda}\lambda^r}{r!} = \frac {e^{-m}m^r}{r!} & \small \color{#3D99F6} \text{where the mean is }\lambda = m = np. \end{aligned}

⟹ A + B + C + D = − m + m + r + r = 7 + 7 = 14 \implies A+B+C+D = -m+m+r+r = 7+7 = \boxed{14}

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