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Question Number 197985 by mokys last updated on 07/Oct/23
lim_(n→∞) ( _( k) ^( n) ) p^k q^(n−k)
$$\underset{{n}\rightarrow\infty} {{lim}}\:\left(\underset{\:{k}} {\overset{\:\:{n}} {\:}}\:\right)\:{p}^{{k}} \:{q}^{{n}−{k}} \\ $$
Commented by mr W last updated on 07/Oct/23
question in current form makes no sense. please check your question.
$${question}\:{in}\:{current}\:{form}\:{makes}\:{no} \\ $$$${sense}. \\ $$$${please}\:{check}\:{your}\:{question}. \\ $$
Commented by mokys last updated on 07/Oct/23
q = 1 − p
$${q}\:=\:\mathrm{1}\:−\:{p} \\ $$
Commented by mr W last updated on 07/Oct/23
and k=?
$${and}\:{k}=? \\ $$
Commented by TheHoneyCat last updated on 08/Oct/23
Well actually, it can make perfect sense, rigorously speaking. (But I understand what you mean... I believe the person who wrote that didn't write what they think they wrote ��)
Commented by TheHoneyCat last updated on 08/Oct/23
I will hereby suppose that (n,k)∈N^2 and (p,q)∈C^2 of course the same problem can be tackled with (n,k)∈(C\N_− )^2 but I bieleve you would not know the meaning (plus, I′m too tired to do complex analysis right now) this gives ((n),(k) ) :=((n!)/(k! (n-k)!)) now, using the folowing formula: n! ∼_(n→∞) (√(2πn))((n/e))^n (n-k)!∼_(n→∞) (√(2π(n−k)))(((n−k)/e))^(n−k) Therefore: ((n),(k) ) p^k q^(n−k) =(p^k /(k!)) q^(n−k) ((n!)/((n−k)!)) ∼((p^k e^k )/(k!q^k )) q^n ((√(2πn))/( (√(2π(n-k))))) (n−k)^k ((n/(n−k)))^n Let′s consider every single block: ((p^k e^k )/(k!q^k )) is a constante. So if p=0 or q=0 the limit is zero (else keep reading). ((√(2πn))/( (√(2π(n-k))))) this thing goes to 1 ((n/(n−k)))^n this goes to e^k (so a constant) now (n−k)^k goes to ∞ (but it′s a polynomial) and q^n goes to 0 exponantialy (if ∣q∣<1) is 1 if ∣q∣=1 and goes to infinity if q∈R and q>1. Otherwise your limit is undefined in C (no phase convergence) So as a conclusion: let l be your limit. p=0 ⇒ l=0 q∈U_(int) :={z∈C: ∣z∣<1} ⇒ l=0 q∈[1,∞[_R ⇒ l=((p^k e^(2k) )/(k!q^k ))∞ (which is probably ∞ in your case...) q∉U_(int) ∪R_+ ⇒ l∈ C ∪ U∞ with U∞:={e^(iθ) ∞, θ∈R}
$$\mathrm{I}\:\mathrm{will}\:\mathrm{hereby}\:\mathrm{suppose}\:\mathrm{that}\:\left({n},{k}\right)\in\mathbb{N}^{\mathrm{2}} \:\mathrm{and} \\ $$$$\left({p},{q}\right)\in\mathbb{C}^{\mathrm{2}} \\ $$$$\mathrm{of}\:\mathrm{course}\:\mathrm{the}\:\mathrm{same}\:\mathrm{problem}\:\mathrm{can}\:\mathrm{be}\:\mathrm{tackled} \\ $$$$\mathrm{with}\:\left({n},{k}\right)\in\left(\mathbb{C}\backslash\mathbb{N}_{−} \right)^{\mathrm{2}} \:\mathrm{but}\:\mathrm{I}\:\mathrm{bieleve}\:\mathrm{you}\:\mathrm{would} \\ $$$$\mathrm{not}\:\mathrm{know}\:\mathrm{the}\:\mathrm{meaning}\:\left(\mathrm{plus},\:\mathrm{I}'\mathrm{m}\:\mathrm{too}\right. \\ $$$$\left.\mathrm{tired}\:\mathrm{to}\:\mathrm{do}\:\mathrm{complex}\:\mathrm{analysis}\:\mathrm{right}\:\mathrm{now}\right) \\ $$$$ \\ $$$$\mathrm{this}\:\mathrm{gives} \\ $$$$\begin{pmatrix}{{n}}\\{{k}}\end{pmatrix}\::=\frac{{n}!}{{k}!\:\left({n}-{k}\right)!} \\ $$$$\mathrm{now},\:\mathrm{using}\:\mathrm{the}\:\mathrm{folowing}\:\mathrm{formula}: \\ $$$${n}!\:\underset{{n}\rightarrow\infty} {\sim}\sqrt{\mathrm{2}\pi{n}}\left(\frac{{n}}{{e}}\right)^{{n}} \\ $$$$\left({n}-{k}\right)!\underset{{n}\rightarrow\infty} {\sim}\sqrt{\mathrm{2}\pi\left({n}−{k}\right)}\left(\frac{{n}−{k}}{{e}}\right)^{{n}−{k}} \\ $$$$ \\ $$$$\mathrm{Therefore}: \\ $$$$\begin{pmatrix}{{n}}\\{{k}}\end{pmatrix}\:{p}^{{k}} \:{q}^{{n}−{k}} \\ $$$$=\frac{{p}^{{k}} }{{k}!}\:\:{q}^{{n}−{k}} \:\frac{{n}!}{\left({n}−{k}\right)!} \\ $$$$\sim\frac{{p}^{{k}} {e}^{{k}} }{{k}!{q}^{{k}} }\:\:{q}^{{n}} \:\frac{\sqrt{\mathrm{2}\pi{n}}}{\:\sqrt{\mathrm{2}\pi\left({n}-{k}\right)}}\:\left({n}−{k}\right)^{{k}} \:\left(\frac{{n}}{{n}−{k}}\right)^{{n}} \\ $$$$ \\ $$$$\mathrm{Let}'\mathrm{s}\:\mathrm{consider}\:\mathrm{every}\:\mathrm{single}\:\mathrm{block}: \\ $$$$\frac{{p}^{{k}} {e}^{{k}} }{{k}!{q}^{{k}} }\:\:\:\mathrm{is}\:\mathrm{a}\:\mathrm{constante}.\:\mathrm{So}\:\mathrm{if}\:{p}=\mathrm{0}\:\mathrm{or}\:{q}=\mathrm{0}\: \\ $$$$\mathrm{the}\:\mathrm{limit}\:\mathrm{is}\:\mathrm{zero}\:\left(\mathrm{else}\:\mathrm{keep}\:\mathrm{reading}\right). \\ $$$$\frac{\sqrt{\mathrm{2}\pi{n}}}{\:\sqrt{\mathrm{2}\pi\left({n}-{k}\right)}}\:\mathrm{this}\:\mathrm{thing}\:\mathrm{goes}\:\mathrm{to}\:\mathrm{1} \\ $$$$\:\left(\frac{{n}}{{n}−{k}}\right)^{{n}} \:\:\:\mathrm{this}\:\mathrm{goes}\:\mathrm{to}\:{e}^{{k}} \:\left(\mathrm{so}\:\mathrm{a}\:\mathrm{constant}\right) \\ $$$$\mathrm{now}\:\left({n}−{k}\right)^{{k}} \:\mathrm{goes}\:\mathrm{to}\:\infty\:\left(\mathrm{but}\:\mathrm{it}'\mathrm{s}\:\mathrm{a}\:\mathrm{polynomial}\right) \\ $$$$\mathrm{and}\:{q}^{{n}} \:\mathrm{goes}\:\mathrm{to}\:\mathrm{0}\:\mathrm{exponantialy}\:\left(\mathrm{if}\:\mid{q}\mid<\mathrm{1}\right) \\ $$$$\mathrm{is}\:\mathrm{1}\:\mathrm{if}\:\mid{q}\mid=\mathrm{1}\:\mathrm{and}\:\:\mathrm{goes}\:\mathrm{to}\:\mathrm{infinity}\:\mathrm{if}\:\:{q}\in\mathbb{R}\:\mathrm{and} \\ $$$${q}>\mathrm{1}.\:\mathrm{Otherwise}\:\mathrm{your}\:\mathrm{limit}\:\mathrm{is}\:\mathrm{undefined}\:\mathrm{in}\:\mathbb{C} \\ $$$$\left({no}\:{phase}\:{convergence}\right) \\ $$$$ \\ $$$$ \\ $$$$\mathrm{So}\:\mathrm{as}\:\mathrm{a}\:\mathrm{conclusion}: \\ $$$$\mathrm{let}\:{l}\:\mathrm{be}\:\mathrm{your}\:\mathrm{limit}. \\ $$$${p}=\mathrm{0}\:\Rightarrow\:{l}=\mathrm{0} \\ $$$${q}\in\mathbb{U}_{\mathrm{int}} :=\left\{{z}\in\mathbb{C}:\:\mid{z}\mid<\mathrm{1}\right\}\:\:\Rightarrow\:{l}=\mathrm{0} \\ $$$${q}\in\left[\mathrm{1},\infty\left[_{\mathbb{R}} \:\Rightarrow\:{l}=\frac{{p}^{{k}} {e}^{\mathrm{2}{k}} }{{k}!{q}^{{k}} }\infty\:\left(\mathrm{which}\:\mathrm{is}\:\mathrm{probably}\:\infty\right.\right.\right. \\ $$$$\left.\mathrm{in}\:\mathrm{your}\:\mathrm{case}…\right) \\ $$$${q}\notin\mathbb{U}_{\mathrm{int}} \cup\mathbb{R}_{+} \Rightarrow\:{l}\in\:\mathbb{C}\:\cup\:\mathbb{U}\infty \\ $$$$\mathrm{with}\:\mathbb{U}\infty:=\left\{{e}^{{i}\theta} \infty,\:\theta\in\mathbb{R}\right\} \\ $$

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