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Question Number 185776 by TUN last updated on 27/Jan/23

	Answered by MJS_new last updated on 27/Jan/23

	$n! \sim \frac{n^{n}}{e^{n}} \sqrt{2 π n}$ $\lim_{n \to \infty} {(\frac{(2 n)!}{n! n^{n}})}^{1 / n} = \lim_{n \to \infty} \frac{4 \times 2^{1 / (2 n)}}{e} = \frac{4}{e}$
	Answered by CElcedricjunior last updated on 27/Jan/23
	$lim_(n→∞) ((((2n)!)/(n!n^n )))^(1/n) =lim_(n→∞) (((Π_(k=1) ^(2n) k)/(Π_(k=1) ^n k×n^n )))^(1/n) =lim_(n→∞) (((Π_(k=n+1) ^(2n) k)/n^n ))^(1/n) =lim_(n→∞) (Π_(k=n+1) ^(2n) ((k/n)))^(1/n) =>en applicant ln lim_(n→∞) (1/n)ln(Π_(k=n+1) ^(2n) ((k/n))) =lim_(n→∞) (1/n)[ln((1+n)/n)+ln((2+n)/n)+.....ln((2n)/n)] =lim_(n→∞) (1/n)Σ_(k=n+1) ^(2n) ln((k/n)) posons { ((p=k−n−1 pour k=n+1=>p=0)),((pour k=2n =>p=n−1)) :} =lim_(n→∞) (1/n)Σ_(p=0) ^(n−1) ln(((p+n+1)/n)) =lim_(n→∞) (1/n)Σ_(p=1) ^n ln(1+(p/n))=lim_(n→∞) ((b−a)/n)Σ_(p=1) ^n f(a+((b−a)/n)p) avec a=1;b=2;f(x)=ln(x) d′apre^� s notre celebre frere Riemann lim_(n→∞) ln((((2n)!)/(n!n^n )))^(1/n) =∫_1 ^2 lnxdx=[xlnx−x]_1 ^2 lim_(n→∞) ((((2n)!)/(n!n^n )))^(1/n) =e^(2ln2−1) =(4/e) ======================= ...............le celebre cedric junior.......... ================== ==$
	$\lim_{n \to \infty} {(\frac{(2 n)!}{n! n^{n}})}^{\frac{1}{n}} = \lim_{n \to \infty} {(\frac{\underset{k = 1}{\prod^{2 n}} k}{\underset{k = 1}{\prod^{n}} k \times n^{n}})}^{\frac{1}{n}}$ $= \lim_{n \to \infty} {(\frac{\underset{k = n + 1}{\prod^{2 n}} k}{n^{n}})}^{\frac{1}{n}} = \lim_{n \to \infty} {(\underset{k = n + 1}{\prod^{2 n}} (\frac{k}{n}))}^{\frac{1}{n}}$ $=> e n a p p l i c a n t l n$ $\lim_{n \to \infty} \frac{1}{n} l n (\underset{k = n + 1}{\prod^{2 n}} (\frac{k}{n}))$ $= \lim_{n \to \infty} \frac{1}{n} [l n \frac{1 + n}{n} + l n \frac{2 + n}{n} + . . . . . l n \frac{2 n}{n}]$ $= \lim_{n \to \infty} \frac{1}{n} \underset{k = n + 1}{\sum^{2 n}} l n (\frac{k}{n})$ $p o s o n s {\begin{cases} p = k - n - 1 p o u r k = n + 1 => p = 0 \\ p o u r k = 2 n => p = n - 1 \end{cases}$ $= \lim_{n \to \infty} \frac{1}{n} \underset{p = 0}{\sum^{n - 1}} l n (\frac{p + n + 1}{n})$ $= \lim_{n \to \infty} \frac{1}{n} \underset{p = 1}{\sum^{n}} l n (1 + \frac{p}{n}) = \lim_{n \to \infty} \frac{b - a}{n} \underset{p = 1}{\sum^{n}} f (a + \frac{b - a}{n} p)$ $a v e c a = 1; b = 2; f (x) = l n (x)$ $d^{'} a p r \overset{`}{e s} n o t r e c e l e b r e f r e r e R i e m a n n$ $\lim_{n \to \infty} l n {(\frac{(2 n)!}{n! n^{n}})}^{\frac{1}{n}} = \int_{1}^{2} l n x d x = {[x l n x - x]}_{1}^{2}$ $\lim_{n \to \infty} {(\frac{(2 n)!}{n! n^{n}})}^{\frac{1}{n}} = e^{2 l n 2 - 1} = \frac{4}{e}$ $=======================$ $. . . . . . . . . . . . . . . l e c e l e b r e c e d r i c j u n i o r . . . . . . . . . .$ $==================$ $==$
	Commented by MJS_new last updated on 27/Jan/23
	�� why use 1 line when you can use 10? ��
	Commented by CElcedricjunior last updated on 28/Jan/23

	$◼ ◼$
	Answered by qaz last updated on 28/Jan/23

	$\underset{n \to \infty}{l i m} {(\frac{(2 n)!}{n! n^{n}})}^{\frac{1}{n}} = \underset{n \to \infty}{l i m} \frac{\frac{(2 n)!}{n! n^{n}}}{\frac{(2 n - 2)!}{(n - 1)! {(n - 1)}^{n - 1}}}$ $= \underset{n \to}{l i m} \frac{2 (2 n - 1)}{n - 1} {(1 - \frac{1}{n})}^{n} = 4 e^{- 1}$
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