lim_(n→∞) n^(−3/2) [(n+1)^((n+1)) (n+2)^((n+2)) ...(2n)^(2n) ]^(1/n^2 ) = ?


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Question Number 204574 by universe last updated on 22/Feb/24

$\lim_{n \to \infty} n^{- 3 / 2} {[{(n + 1)}^{(n + 1)} {(n + 2)}^{(n + 2)} . . . {(2 n)}^{2 n}]}^{1 / n^{2}} = ?$
	Answered by TonyCWX08 last updated on 22/Feb/24

	$k n o w$ $\lim_{n \to \infty} (\frac{1}{n^{a}}) = 0, a = P o s i t i v e N u m b e r$ $n^{- \frac{3}{2}} = \frac{1}{n^{\frac{3}{2}}}$ $\lim_{n \to \infty} (\frac{1}{n^{\frac{3}{2}}})$ $= 0$ $A n y t h i n g M u l t i p l y 0 i s 0$ $F i n a l a n s w e r = 0$
	Commented by mr W last updated on 22/Feb/24

	$A n y t h i n g M u l t i p l y 0 i s 0 ?$ $(2 n^{2} + 1) \times \frac{1}{n^{2}} \to 2$
	Commented by TonyCWX08 last updated on 22/Feb/24

	$T h e z e r o c a m e f r o m t h e l i m i t o f n^{- 3 / 2}$ $U n d e r s t a n d ?$
	Commented by mr W last updated on 22/Feb/24

	$i j u s t w a n t e d t o g i v e a n e x a m p l e$ $t h a t \infty \times 0 \neq 0 . s o i t i s n o t e n o u g h o n l y$ $t o k n o w n^{- \frac{3}{2}} \to 0 a n d t h e n s a y t h e$ $f i n a l l i m i t i s z e r o . w h a t i s a b o u t$ ${[{(n + 1)}^{(n + 1)} . . . . {(2 n)}^{2 n}]}^{\frac{1}{n^{2}}} ?$
	Answered by witcher3 last updated on 22/Feb/24

	$Ω = \frac{1}{n \sqrt{n}} e^{\frac{1}{n^{2}} [\underset{k = 1}{\sum^{n}} (n + k) \ln (n + k)]} = \frac{1}{n \sqrt{n}} e^{\frac{1}{n} (\underset{k = 1}{\sum^{n}} (1 + \frac{k}{n}) (\ln (n) + \ln (1 + \frac{k}{n}))}$ $= \frac{e^{\frac{1}{n} \underset{k = 1}{\sum^{n}} (1 + \frac{k}{n}) \ln (n) + \frac{1}{n} \underset{k = 1}{\sum^{n}} (1 + \frac{k}{n}) \ln (1 + \frac{k}{n})}}{n \sqrt{n}}$ $= \frac{\ln (n)}{n} \underset{k = 1}{\sum^{n}} (1 + \frac{k}{n}) = \frac{\ln (n)}{n} (n + \frac{n + 1}{2}) . . S_{1}$ $\lim_{n \to 0} \frac{1}{n} \underset{1}{\sum^{n}} (1 + \frac{k}{n}) \ln (1 + \frac{k}{n}) \overset{Riemann Sum}{=} \int_{0}^{1} (1 + x) \ln (1 + x) dx$ $\int_{0}^{1} (1 + x) \ln (1 + x) dx = {[\frac{{(1 + x)}^{2}}{2} \ln (1 + x)]}_{0}^{1} - \int_{0}^{1} \frac{1 + x}{2} dx = 2 \ln (2) - \frac{3}{2}$ $\Rightarrow \forall ϵ > 0 \exists N \in N; \forall n ⩾ N; 2 \ln (2) - \frac{3}{2} - ϵ ⩽ S_{2} = \frac{1}{n} \underset{k = 1}{\sum^{n}} (1 + \frac{k}{n}) \ln (1 + \frac{k}{n}) ⩽ 2 \ln (2) - \frac{3}{2} + ϵ . . .$ $Ω = \frac{1}{n \sqrt{n}} e^{\frac{1}{n} (\frac{3 n + 1}{2}) \ln (n) + S_{2}} = \frac{e^{\frac{3}{2} \ln (n) + \frac{\ln (n)}{2 n} + S_{2}}}{n \sqrt{n}} = \frac{n^{\frac{3}{2}}}{n^{\frac{3}{2}}} . e^{\frac{\ln (n)}{2 n} + S_{2}}$ $= e^{\frac{\ln (n)}{2 n} + S_{2}}; \frac{\ln (n)}{2 n} \to 0; S_{2} \to 2 \ln (2) - \frac{3}{2}$ $\lim_{n \to \infty} Ω = e^{2 \ln (2) - \frac{3}{2}} = \frac{4}{e \sqrt{e}}$
	Commented by universe last updated on 22/Feb/24

	$t h a n k u s o m u c h s i r$
	Commented by universe last updated on 22/Feb/24

	$s i r \int_{0}^{1} (1 + x) \log (1 + x) d x = 2 l o g 2 - 3 / 4$
	Commented by witcher3 last updated on 22/Feb/24

	$yes sorry for the mistack Too busy$
	Commented by universe last updated on 23/Feb/24

	$n o p r o b l e m s i r y o u r a p p r o c h i s r i g h t$
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