Π_(k=1) ^n (1+(k^2 /n^2 ))^(1/n)


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Question Number 150263 by dany last updated on 10/Aug/21

$\underset{k = 1}{\prod^{n}} {(1 + \frac{k^{2}}{n^{2}})}^{\frac{1}{n}}$
Commented by puissant last updated on 10/Aug/21

$limite ?$
Commented by dany last updated on 10/Aug/21

$y e s$
	Answered by puissant last updated on 11/Aug/21
	$U_n =Π_(k=1) ^n (1+(k^2 /n^2 ))^(1/n) let V_n =ln(U_n ) lim_(n→+∞) V_n =lim_(n→+∞) (1/n)Σ_(k=1) ^n ln(1+((k/n))^2 ) (Riemann integral ((b−a)/n)Σ_(k=1) ^n f(a+k((b−a)/n))) we have lim_(n→+∞) V_n = ∫_0 ^1 ln(1+x^2 )dx K=∫_0 ^1 ln(1+x^2 )dx { ((u=ln(1+x^2 ))),((v′=1)) :}⇒ { ((u′=((2x)/(1+x^2 )))),((v=x)) :} K=[xln(1+x^2 )]_0 ^1 −2∫_0 ^1 (x^2 /(1+x^2 ))dx =ln2−2∫_0 ^1 ((x^2 +1)/(x^2 +1))dx+2∫_0 ^1 (1/(1+x^2 ))dx =ln2−2[x]_0 ^1 +2[arctan(x)]_0 ^1 =ln2−2+(π/2).. lim_(n→+∞) V_n =lim_(n→+∞) ln(U_n ) ⇒ lim_(n→+∞) U_n = e^(lim_(n→+∞) V_n ) ⇒ lim_(n→+∞) U_n = e^(ln2−2+(π/2)) lim_(n→+∞) Π_(k=1) ^n (1+(k^2 /n^2 ))^(1/n) = 2e^((π/2)−2) .. ...........Le puissant...........$
	$U_{n} = \underset{k = 1}{\prod^{n}} {(1 + \frac{k^{2}}{n^{2}})}^{\frac{1}{n}} let V_{n} = \ln (U_{n})$ $\lim_{n \to + \infty} V_{n} = \lim_{n \to + \infty} \frac{1}{n} \underset{k = 1}{\sum^{n}} \ln (1 + {(\frac{k}{n})}^{2})$ $(Riemann integral \frac{b - a}{n} \underset{k = 1}{\sum^{n}} f (a + k \frac{b - a}{n}))$ $we have \lim_{n \to + \infty} V_{n} = \int_{0}^{1} \ln (1 + x^{2}) dx$ $K = \int_{0}^{1} \ln (1 + x^{2}) dx$ ${\begin{cases} u = \ln (1 + x^{2}) \\ v^{'} = 1 \end{cases} \Rightarrow {\begin{cases} u^{'} = \frac{2 x}{1 + x^{2}} \\ v = x \end{cases}$ $K = {[xln (1 + x^{2})]}_{0}^{1} - 2 \int_{0}^{1} \frac{x^{2}}{1 + x^{2}} dx$ $= \ln 2 - 2 \int_{0}^{1} \frac{x^{2} + 1}{x^{2} + 1} dx + 2 \int_{0}^{1} \frac{1}{1 + x^{2}} dx$ $= \ln 2 - 2 {[x]}_{0}^{1} + 2 {[\arctan (x)]}_{0}^{1}$ $= \ln 2 - 2 + \frac{π}{2} . .$ $\lim_{n \to + \infty} V_{n} = \lim_{n \to + \infty} \ln (U_{n})$ $\Rightarrow \lim_{n \to + \infty} U_{n} = e^{\lim_{n \to + \infty} V_{n}}$ $\Rightarrow \lim_{n \to + \infty} U_{n} = e^{\ln 2 - 2 + \frac{π}{2}}$ $\lim_{n \to + \infty} \underset{k = 1}{\prod^{n}} {(1 + \frac{k^{2}}{n^{2}})}^{\frac{1}{n}} = {2 e}^{\frac{π}{2} - 2} . .$ $. . . . . . . . . . . Le puissant . . . . . . . . . . .$
	Answered by Ar Brandon last updated on 10/Aug/21

	$L = \lim_{n \to \infty} \underset{k = 1}{\prod^{n}} {(1 + \frac{k^{2}}{n^{2}})}^{\frac{1}{n}}$ $\ln L = \lim_{n \to \infty} \frac{1}{n} \ln \underset{k = 1}{\prod^{n}} (1 + \frac{k^{2}}{n^{2}}) = \lim_{n \to \infty} \frac{1}{n} \underset{k = 1}{\sum^{n}} \ln (1 + \frac{k^{2}}{n^{2}})$ $= \int_{0}^{1} \ln (1 + x^{2}) d x = {[x \ln (1 + x^{2})]}_{0}^{1} - 2 \int_{0}^{1} \frac{x^{2}}{1 + x^{2}} d x$ $= \ln 2 - 2 \int_{0}^{1} (1 - \frac{1}{1 + x^{2}}) d x = \ln 2 - 2 + \frac{π}{2}$ $L = e^{\ln 2 - 2 + \frac{π}{2}} = 2 e^{\frac{π}{2} - 2}$
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