W_n =Σ_(k=1) ^(2n) (k/(n^2 +k^2 )) montrer que W_n converge et calculer la valeur de W


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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
	$W_n =Σ_(k=1) ^n (k/(n^2 +k^2 )) +Σ_(k=n+1) ^(2n) (k/(n^2 +k^2 )) =u_n +v_n u_n =Σ_(k=1) ^n ((k/n^2 )/(1+(k^2 /n^2 ))) =(1/n)Σ_(k=1) ^n ((k/n)/(1+((k/n))^2 ))→∫_0 ^1 (x/(1+x^2 ))dx =(1/2)∫_0 ^1 ((2x)/(1+x^2 ))dx =(1/2)[ln(1+x^2 )]_0 ^1 =((ln(2))/2) v_n =Σ_(k=n+1) ^(2n) (k/(n^2 +k^2 )) =_(k−n=p) Σ_(p=1) ^n ((n+p)/(n^2 +(n+p)^2 )) =Σ_(p=1) ^n (((n+p)/n^2 )/(1+(((n+p)/n))^2 )) =(1/n)Σ_(p=1) ^n ((1+(p/n))/(1+(1+(p/n))^2 )) →∫_0 ^1 ((1+x)/(1+(1+x)^2 ))dx =_(1+x=t) ∫_1 ^2 (t/(1+t^2 ))dt =(1/2)[ln(1+t^2 )]_1 ^2 =(1/2){ln(5)−ln(2)} ⇒lim_(n→+∞) W_n =((ln2)/2) +((ln(5))/2)−((ln(2))/2) ⇒ lim_(n→+∞) =(1/2)ln(5)$
	$W_{n} = \sum_{k = 1}^{n} \frac{k}{n^{2} + k^{2}} + \sum_{k = n + 1}^{2 n} \frac{k}{n^{2} + k^{2}} = u_{n} + v_{n}$ $u_{n} = \sum_{k = 1}^{n} \frac{\frac{k}{n^{2}}}{1 + \frac{k^{2}}{n^{2}}} = \frac{1}{n} \sum_{k = 1}^{n} \frac{\frac{k}{n}}{1 + {(\frac{k}{n})}^{2}} \to \int_{0}^{1} \frac{x}{1 + x^{2}} dx = \frac{1}{2} \int_{0}^{1} \frac{2 x}{1 + x^{2}} dx$ $= \frac{1}{2} {[\ln (1 + x^{2})]}_{0}^{1} = \frac{\ln (2)}{2}$ $v_{n} = \sum_{k = n + 1}^{2 n} \frac{k}{n^{2} + k^{2}} =_{k - n = p} \sum_{p = 1}^{n} \frac{n + p}{n^{2} + {(n + p)}^{2}}$ $= \sum_{p = 1}^{n} \frac{\frac{n + p}{n^{2}}}{1 + {(\frac{n + p}{n})}^{2}} = \frac{1}{n} \sum_{p = 1}^{n} \frac{1 + \frac{p}{n}}{1 + {(1 + \frac{p}{n})}^{2}}$ $\to \int_{0}^{1} \frac{1 + x}{1 + {(1 + x)}^{2}} dx =_{1 + x = t} \int_{1}^{2} \frac{t}{1 + t^{2}} dt = \frac{1}{2} {[\ln (1 + t^{2})]}_{1}^{2}$ $= \frac{1}{2} {\ln (5) - \ln (2)} \Rightarrow \lim_{n \to + \infty} W_{n} = \frac{\ln 2}{2} + \frac{\ln (5)}{2} - \frac{\ln (2)}{2} \Rightarrow$ $\lim_{n \to + \infty} = \frac{1}{2} \ln (5)$
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