W-n-k-1-2n-k-n-2-k-2-montrer-que-W-n-converge-et-calculer-la-valeur-de-W-n-

Question Number 123826 by pticantor last updated on 28/Nov/20

W_{n} = \underset{k = 1}{\sum^{2 n}} \frac{k}{n^{2} + k^{2}}

m o n t r e r q u e W_{n} c o n v e r g e

e t c a l c u l e r l a v a l e u r d e W_{n}

Commented by Dwaipayan Shikari last updated on 28/Nov/20

W_{n} = \lim_{n \to \infty} \underset{k = 1}{\sum^{2 n}} \frac{k}{n^{2} + k^{2}}

\frac{1}{n} \underset{k = 1}{\sum^{2 n}} \frac{\frac{k}{n}}{1 + {(\frac{k}{n})}^{2}} = \int_{0}^{2} \frac{x}{1 + x^{2}} d x = \frac{1}{2} {[l o g (1 + x^{2})]}_{0}^{2} = l o g (\sqrt{5})

Answered by mathmax by abdo last updated on 28/Nov/20

W_{n} = \sum_{k = 1}^{n} \frac{k}{n^{2} + k^{2}} + \sum_{k = n + 1}^{2 n} \frac{k}{n^{2} + k^{2}} = u_{n} + v_{n}

u_{n} = \sum_{k = 1}^{n} \frac{\frac{k}{n^{2}}}{1 + \frac{k^{2}}{n^{2}}} = \frac{1}{n} \sum_{k = 1}^{n} \frac{\frac{k}{n}}{1 + {(\frac{k}{n})}^{2}} \to \int_{0}^{1} \frac{x}{1 + x^{2}} dx = \frac{1}{2} \int_{0}^{1} \frac{2 x}{1 + x^{2}} dx

= \frac{1}{2} {[\ln (1 + x^{2})]}_{0}^{1} = \frac{\ln (2)}{2}

v_{n} = \sum_{k = n + 1}^{2 n} \frac{k}{n^{2} + k^{2}} =_{k - n = p} \sum_{p = 1}^{n} \frac{n + p}{n^{2} + {(n + p)}^{2}}

= \sum_{p = 1}^{n} \frac{\frac{n + p}{n^{2}}}{1 + {(\frac{n + p}{n})}^{2}} = \frac{1}{n} \sum_{p = 1}^{n} \frac{1 + \frac{p}{n}}{1 + {(1 + \frac{p}{n})}^{2}}

\to \int_{0}^{1} \frac{1 + x}{1 + {(1 + x)}^{2}} dx =_{1 + x = t} \int_{1}^{2} \frac{t}{1 + t^{2}} dt = \frac{1}{2} {[\ln (1 + t^{2})]}_{1}^{2}

= \frac{1}{2} {\ln (5) - \ln (2)} \Rightarrow \lim_{n \to + \infty} W_{n} = \frac{\ln 2}{2} + \frac{\ln (5)}{2} - \frac{\ln (2)}{2} \Rightarrow

\lim_{n \to + \infty} = \frac{1}{2} \ln (5)