let-u-n-1-n-2-k-1-n-n-2-k-2-1-n-determine-lim-n-u-n-

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} = \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