Ω_n = determinant ((1,1,1,(...),1),(1,2^2 ,2^3 ,(...),2^n ),(1,3^2 ,3^3 ,(...),3^n ),((...),(...),(...),(...),(...)),(1,n^2 ,n^3 ,(...),n^n )) , n ∈ N^∗ Find: Ω =lim_(n→∞) ((Ω_(n+1) /Ω


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Question Number 181183 by Shrinava last updated on 22/Nov/22

$Ω_{n} = \| \begin{matrix} 1 & 1 & 1 & . . . & 1 \\ 1 & 2^{2} & 2^{3} & . . . & 2^{n} \\ 1 & 3^{2} & 3^{3} & . . . & 3^{n} \\ . . . & . . . & . . . & . . . & . . . \\ 1 & n^{2} & n^{3} & . . . & n^{n} \end{matrix} \|, n \in N^{*}$ $Find : Ω = \lim_{n \to \infty} \sqrt[n]{\frac{Ω_{n + 1}}{Ω_{n}}}$
	Answered by aleks041103 last updated on 22/Nov/22

	$t h e d e t e r m i a n t D (x_{i}) :$ $D (x_{i}) = \| \begin{matrix} 1 & 1 & . . . & 1 \\ x_{1}^{2} & x_{2}^{2} & . . . & x_{n}^{2} \\ x_{1}^{3} & x_{2}^{3} & . . . & x_{n}^{3} \\ . . . & . . . & . . . & . . . \\ x_{1}^{n} & x_{2}^{n} & . . . & x_{n}^{n} \end{matrix} \|$ $b y l a p l a c e f o r m u l a$ $D = ϵ_{i_{1} i_{2} . . . i_{n}} x_{i_{1}}^{0} x_{i_{2}}^{2} x_{i_{3}}^{3} . . . x_{i_{n}}^{n} =$ $= ϵ_{i_{1} . . . i_{n}} x_{i_{2}}^{2} . . . x_{i_{n}}^{2} (x_{i_{2}}^{0} x_{i_{3}}^{1} . . . x_{i_{n}}^{n - 2}) =$ $= {(x_{1} x_{2} . . . x_{n})}^{2} ϵ_{i_{1} . . . i_{n}} \frac{1}{x_{i_{1}}^{2}} (x_{i_{2}}^{0} x_{i_{3}}^{1} . . . x_{i_{n}}^{n - 2})$ $ϵ_{k σ (12 . . . (k - 1) (k + 1) . . . n)} = {(- 1)}^{s g n (σ)} ϵ_{k 12 . . . (k - 1) (k + 1) . . . n} =$ $= {(- 1)}^{s g n (σ)} {(- 1)}^{k - 1} ϵ_{12 . . . (k - 1) k (k + 1) . . . n} =$ $= {(- 1)}^{k - 1} ϵ_{σ (12 . . (k - 1) (k + 1) . . n)}$ $\Rightarrow ϵ_{i_{1} . . . i_{n}} = {(- 1)}^{i_{1} - 1} ϵ_{i_{2} . . . i_{n}}$ $\Rightarrow D = {(x_{1} x_{2} . . . x_{n})}^{2} \underset{s = 1}{\sum^{n}} \frac{{(- 1)}^{s - 1}}{x_{s}^{2}} {(ϵ_{i_{2} . . . i_{n}} x_{i_{2}}^{0} x_{i_{3}}^{1} . . . x_{i_{n}}^{n - 2})}_{i_{j} \neq s}$ ${(ϵ_{i_{2} . . . i_{n}} x_{i_{2}}^{0} x_{i_{3}}^{1} . . . x_{i_{n}}^{n - 2})}_{i_{j} \neq s} = \| \begin{matrix} 1 & 1 & . . . & 1 & 1 & . . . & 1 \\ x_{1} & x_{2} & . . . & x_{s - 1} & x_{s + 1} & . . . & x_{n} \\ . . . & . . . & . . . & . . . & . . . & . . . & . . . \\ x_{1}^{n - 2} & x_{2}^{n - 2} & . . . & x_{s - 1}^{n - 2} & x_{s + 1}^{n - 2} & . . . & x_{n}^{n - 2} \end{matrix} \| =$ $= \prod_{i > j (i, j \neq s)} (x_{i} - x_{j})$ $\Rightarrow D = {(x_{1} x_{2} . . . x_{n})}^{2} \underset{s = 1}{\sum^{n}} \frac{{(- 1)}^{s - 1}}{x_{s}^{2}} \prod_{i > j (i, j \neq s)} (x_{i} - x_{j})$ $I n o u r c a s e : x_{i} = i$ $\Rightarrow Ω_{n} = {(n!)}^{2} \underset{s = 1}{\sum^{n}} \frac{{(- 1)}^{s - 1}}{s^{2}} \prod_{i > j (i, j \neq s)} (i - j)$ $\prod_{i > j (i, j \neq s)} (i - j) = \frac{\prod_{i > j} (i - j)}{(\prod_{s > j} (s - j)) (\prod_{i > s} (i - s))}$ $\prod_{s > j} (s - j) = \underset{j = 1}{\prod^{s - 1}} (s - j) = (s - 1) (s - 2) . . . 2.1 = (s - 1)!$ $\prod_{i > s} (i - s) = \underset{i = s + 1}{\prod^{n}} (i - s) = 1.2 . . . (n - s) = (n - s)!$ $\prod_{i > j} (i - j) = \underset{j = 1}{\prod^{n - 1}} \underset{i = j + 1}{\prod^{n}} (i - j) =$ $= \underset{j = 1}{\prod^{n - 1}} \underset{i = 1}{\prod^{n - j}} i = \underset{j = 1}{\prod^{n - 1}} (n - j)! = \underset{j = 1}{\prod^{n - 1}} j!$ $\Rightarrow Ω_{n} = (\underset{j = 1}{\prod^{n - 1}} j!) {(n!)}^{2} \underset{s = 1}{\sum^{n}} \frac{{(- 1)}^{s - 1}}{s^{2}} \frac{1}{(s - 1)! (n - s)!} =$ $= (\underset{j = 1}{\prod^{n}} j!) \underset{s = 1}{\sum^{n}} (\begin{matrix} n \\ s \end{matrix}) \frac{{(- 1)}^{s - 1}}{s}$ ${(1 - x)}^{n} = \underset{s = 0}{\sum^{n}} (\begin{matrix} n \\ s \end{matrix}) {(- 1)}^{s} x^{s} = 1 - x \underset{s = 1}{\sum^{n}} (\begin{matrix} n \\ s \end{matrix}) {(- 1)}^{s - 1} x^{s - 1}$ $\Rightarrow \underset{s = 1}{\sum^{n}} (\begin{matrix} n \\ s \end{matrix}) {(- 1)}^{s - 1} x^{s - 1} = \frac{1 - {(1 - x)}^{n}}{x}$ $\Rightarrow \int_{0}^{1} (\underset{s = 1}{\sum^{n}} (\begin{matrix} n \\ s \end{matrix}) {(- 1)}^{s - 1} x^{s - 1}) d x =$ $= \underset{s = 1}{\sum^{n}} (\begin{matrix} n \\ s \end{matrix}) {(- 1)}^{s - 1} (\int_{0}^{1} x^{s - 1} d x) =$ $= \underset{s = 1}{\sum^{n}} (\begin{matrix} n \\ s \end{matrix}) \frac{{(- 1)}^{s}}{s} = \int_{0}^{1} \frac{1 - {(1 - x)}^{n}}{x} d x$ $\int_{0}^{1} \frac{1 - {(1 - x)}^{n}}{x} d x = \int_{0}^{1} \frac{1 - x^{n}}{1 - x} d x = \int_{0}^{1} \underset{k = 0}{\sum^{n - 1}} x^{k} d x =$ $= \underset{k = 0}{\sum^{n - 1}} \int_{0}^{1} x^{k} d x = \underset{k = 0}{\sum^{n - 1}} \frac{1}{k + 1} = \underset{k = 1}{\sum^{n}} \frac{1}{k} = H (n)$ $\Rightarrow Ω_{n} = (1! 2! . . . n!) H (n)$ $\frac{Ω_{n + 1}}{Ω_{n}} = a_{n} = \frac{(1! 2! . . . n! (n + 1)!) H (n + 1)}{(1! 2! . . . n!) H (n)} = (n + 1)! \frac{H (n + 1)}{H (n)}$ $L = \underset{n \to \infty}{l i m} \sqrt[n]{a_{n}} = \underset{n \to \infty}{l i m} \frac{a_{n + 1}}{a_{n}} =$ $= \underset{n \to \infty}{l i m} \frac{(n + 2)! H (n + 2) H (n)}{(n + 1)! {(H (n + 1))}^{2}} = \underset{n \to \infty}{l i m} (n + 2) \frac{H (n) H (n + 2)}{{(H (n + 1))}^{2}}$ $H (n) \sim l n (n)$ $\Rightarrow L = \underset{n \to \infty}{l i m} \frac{(n + 2) l n (n) l n (n + 2)}{l n (n + 1) l n (n + 1)}$ $b u t \underset{n \to \infty}{l i m} \frac{l n (n)}{l n (n + 1)} = 1 \Rightarrow$ $L = \underset{n \to \infty}{l i m} (n + 2) = \infty$ $\Rightarrow \underset{n \to \infty}{l i m} \sqrt[n]{\frac{Ω_{n + 1}}{Ω_{n}}} \to \infty$
	Commented by Shrinava last updated on 23/Nov/22

	$Thank you so much my dear professor,$ $perfect solution as always$
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