let u_n =(1/n^2 )Π_(k=1) ^n (n^2 +k^2 )^(1/n) determine lim_(n→+∞) u


Question and Answers Forum

All Questions Topic List
Relation and Functions Questions
Previous in All Question Next in All Question
Previous in Relation and Functions Next in Relation and Functions
Question Number 107288 by mathmax by abdo last updated on 09/Aug/20

$let u_{n} = \frac{1}{n^{2}} \prod_{k = 1}^{n} {(n^{2} + k^{2})}^{\frac{1}{n}}$ $determine \lim_{n \to + \infty} u_{n}$
	Answered by Ar Brandon last updated on 13/Aug/20
	$u_n =(1/n^2 )Π_(k=1) ^n (n^2 +k^2 )^(1/n) ln(u_n )=−2ln(n)+lnΠ_(k=1) ^n (n^2 +k^2 )^(1/n) =−2ln(n)+(1/n)Σ_(k=1) ^n ln(n^2 +k^2 ) =−2ln(n)+(1/n)Σ_(k=1) ^n [2ln(n)+ln(1+(k^2 /n^2 ))] =−2ln(n)+2ln(n)+(1/n)Σ_(k=1) ^n ln(1+(k^2 /n^2 )) lim_(n→∞) ln(u_n )=lim_(n→∞) (1/n)Σ_(k=1) ^n ln(1+(k^2 /n^2 )) =∫_0 ^1 ln(1+x^2 )dx =[ln(1+x^2 )∫dx−∫{((d ln(1+x^2 ))/dx)∙∫dx}dx]_0 ^1 =[xln(1+x^2 )−∫((2x^2 )/(1+x^2 ))dx]_0 ^1 =ln2−2∫_0 ^1 {1−(1/(1+x^2 ))}dx =ln2−2+(π/2) ⇒lim_(n→∞) u_n =e^(ln2−2+(π/2)) =2e^((π/2)−2)$
	$u_{n} = \frac{1}{n^{2}} \underset{k = 1}{\prod^{n}} {(n^{2} + k^{2})}^{\frac{1}{n}}$ $\ln (u_{n}) = - 2 \ln (n) + \ln \underset{k = 1}{\prod^{n}} {(n^{2} + k^{2})}^{1 / n}$ $= - 2 \ln (n) + \frac{1}{n} \underset{k = 1}{\sum^{n}} \ln (n^{2} + k^{2})$ $= - 2 \ln (n) + \frac{1}{n} \underset{k = 1}{\sum^{n}} [2 \ln (n) + \ln (1 + \frac{k^{2}}{n^{2}})]$ $= - 2 \ln (n) + 2 \ln (n) + \frac{1}{n} \underset{k = 1}{\sum^{n}} \ln (1 + \frac{k^{2}}{n^{2}})$ $\underset{n \to \infty}{\lim l n} (u_{n}) = \lim_{n \to \infty} \frac{1}{n} \underset{k = 1}{\sum^{n}} \ln (1 + \frac{k^{2}}{n^{2}})$ $= \int_{0}^{1} \ln (1 + x^{2}) dx$ $= {[\ln (1 + x^{2}) \int dx - \int {\frac{d \ln (1 + x^{2})}{dx} \cdot \int dx} dx]}_{0}^{1}$ $= {[xln (1 + x^{2}) - \int \frac{{2 x}^{2}}{1 + x^{2}} dx]}_{0}^{1} = \ln 2 - 2 \int_{0}^{1} {1 - \frac{1}{1 + x^{2}}} dx$ $= \ln 2 - 2 + \frac{π}{2}$ $Double subscripts: use braces to clarify$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum