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  "body": "\n## Proof\n\nWe split the binomial probability into three factors and find the limit of each one.\n\n1. Fix an integer $k \\ge 0$. Since $n p_n \\to \\lambda$, we have $p_n = (n p_n)/n \\to \\lambda \\cdot 0 = 0$. So there is $n_0 \\ge k$ with $p_n \\le 1/2$ for $n \\ge n_0$; from now on $n \\ge n_0$.\n2. By the definition of the [[binomial distribution|0VRWZDWW6B]] and of the [[binomial coefficient|092GEF38YD]],\n$$P(S_n = k) = \\binom{n}{k} p_n^k (1 - p_n)^{n-k} = a_n \\cdot \\frac{(n p_n)^k}{k!} \\cdot (1 - p_n)^{n} \\cdot (1 - p_n)^{-k},$$\nwith $a_n = \\frac{n(n-1)\\cdots(n-k+1)}{n^k} = \\prod_{j=0}^{k-1} \\left(1 - \\frac{j}{n}\\right)$.\n3. Each of the $k$ factors $1 - j/n$ tends to $1$. So $a_n \\to 1$, and for $k = 0$ the empty product is $1$.\n4. Since $n p_n \\to \\lambda$, $(n p_n)^k / k! \\to \\lambda^k / k!$. Since $p_n \\to 0$, $(1 - p_n)^{-k} \\to 1$.\n5. Claim: $-x - x^2 \\le \\ln(1 - x) \\le -x$ for $0 \\le x \\le 1/2$. The upper bound is $1 - x \\le e^{-x}$, the inequality [[$1 + y \\le e^y$|012C7GFF20]] with $y = -x$.\n6. For the lower bound, let $g(x) = \\ln(1 - x) + x + x^2$. Then $g(0) = 0$ and $g'(x) = -\\frac{1}{1-x} + 1 + 2x = \\frac{x(1 - 2x)}{1 - x} \\ge 0$ on $[0, 1/2]$. So $g \\ge 0$ there.\n7. By step 5 with $x = p_n$, $-n p_n - n p_n \\cdot p_n \\le n \\ln(1 - p_n) \\le -n p_n$.\n8. Both bounds tend to $-\\lambda$, since $n p_n \\to \\lambda$ and $n p_n \\cdot p_n \\to \\lambda \\cdot 0 = 0$. So $n \\ln(1 - p_n) \\to -\\lambda$.\n9. The exponential is continuous and is the [[inverse of the logarithm|0RKW2X63YR]]. So $(1 - p_n)^n = \\exp(n \\ln(1 - p_n)) \\to e^{-\\lambda}$.\n10. By steps 2, 3, 4 and 9, $P(S_n = k) \\to 1 \\cdot \\frac{\\lambda^k}{k!} \\cdot e^{-\\lambda} \\cdot 1 = e^{-\\lambda} \\lambda^k / k!$. This is the first claim.\n11. Let $T \\sim \\operatorname{Poisson}(\\lambda)$, with the [[Poisson distribution|0VEFSSKX18]]. Let $F_n$ and $F$ be the [[cumulative distribution functions|0MXMS3AE13]] of $S_n$ and $T$.\n12. Both $S_n$ and $T$ take values in $\\set{0, 1, 2, \\dots}$. So for $x \u003c 0$, $F_n(x) = 0 = F(x)$.\n13. For $x \\ge 0$, let $m = \\floor{x}$. Then $F_n(x) = \\sum_{k=0}^{m} P(S_n = k)$ and $F(x) = \\sum_{k=0}^{m} e^{-\\lambda} \\lambda^k / k!$. These are finite sums, so step 10 gives $F_n(x) \\to F(x)$.\n14. Thus $F_n(x) \\to F(x)$ for every real $x$, in particular at every continuity point of $F$. So $S_n \\to T$ [[in distribution|0JDAQHWA9N]].\n\n## Idea\n\nFor fixed $k$, choosing $k$ successes among $n$ trials gives about $n^k/k!$ choices, each with probability about $p_n^k$, so this part tends to $\\lambda^k/k!$. The chance of no other success, $(1 - p_n)^{n}$, tends to $e^{-\\lambda}$ because $\\ln(1 - p) \\approx -p$ for small $p$.\n",
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