Question Number 172307 by mathocean1 last updated on 25/Jun/22
$${Calculate}\: \\ $$$$\underset{{n}\rightarrow+\infty} {{lim}A}_{{n}} =\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{{n}} }{\mathrm{1}+{x}}{dx} \\ $$
Answered by aleks041103 last updated on 25/Jun/22
![A_n =∫_0 ^1 (x^n /(1−(−x)))dx=(−1)^n ∫_0 ^1 (((−x)^n −1+1)/(1−(−x)))dx= =(−1)^(n+1) ∫_0 ^1 ((1−(−x)^n −1)/(1−(−x)))dx= =(−1)^(n+1) [∫_0 ^1 ((1−(−x)^n )/(1−(−x)))dx−∫_0 ^1 (dx/(1+x))]= =(−1)^(n+1) [∫_0 ^1 (Σ_(k=1) ^n (−x)^(k−1) )dx−ln(1+1)+ln(1+0)]= =(−1)^(n+1) [Σ_(k=1) ^n (−1)^(k−1) (∫_0 ^1 x^(k−1) dx)−ln(2)]= =(−1)^(n+1) [Σ_(k=1) ^n (((−1)^(k−1) )/k)−ln(2)] (1/(1+x))=Σ_(k=0) ^∞ (−x)^k ⇒∫_0 ^s (dx/(1+x))=ln(1+s)=Σ_(k=0) ^∞ (−1)^k (s^(k+1) /(k+1))= ⇒ln(1+s)=Σ_(k=1) ^∞ (((−1)^(k−1) s^k )/k) ⇒Σ_(k=1) ^n (((−1)^(k−1) )/k)=ln(2)−Σ_(k=n+1) ^∞ (((−1)^(k−1) )/k) ⇒A_n =(−1)^n Σ_(k=n+1) ^∞ (((−1)^(k−1) )/k) ⇒lim_(n→∞) A_n =0](https://www.tinkutara.com/question/Q172318.png)
$${A}_{{n}} =\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{{n}} }{\mathrm{1}−\left(−{x}\right)}{dx}=\left(−\mathrm{1}\right)^{{n}} \int_{\mathrm{0}} ^{\mathrm{1}} \frac{\left(−{x}\right)^{{n}} −\mathrm{1}+\mathrm{1}}{\mathrm{1}−\left(−{x}\right)}{dx}= \\ $$$$=\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} \int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}−\left(−{x}\right)^{{n}} −\mathrm{1}}{\mathrm{1}−\left(−{x}\right)}{dx}= \\ $$$$=\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} \left[\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}−\left(−{x}\right)^{{n}} }{\mathrm{1}−\left(−{x}\right)}{dx}−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dx}}{\mathrm{1}+{x}}\right]= \\ $$$$=\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} \left[\int_{\mathrm{0}} ^{\mathrm{1}} \left(\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left(−{x}\right)^{{k}−\mathrm{1}} \right){dx}−{ln}\left(\mathrm{1}+\mathrm{1}\right)+{ln}\left(\mathrm{1}+\mathrm{0}\right)\right]= \\ $$$$=\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} \left[\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} \left(\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{k}−\mathrm{1}} {dx}\right)−{ln}\left(\mathrm{2}\right)\right]= \\ $$$$=\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} \left[\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} }{{k}}−{ln}\left(\mathrm{2}\right)\right] \\ $$$$\frac{\mathrm{1}}{\mathrm{1}+{x}}=\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−{x}\right)^{{k}} \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{{s}} \frac{{dx}}{\mathrm{1}+{x}}={ln}\left(\mathrm{1}+{s}\right)=\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{k}} \frac{{s}^{{k}+\mathrm{1}} }{{k}+\mathrm{1}}= \\ $$$$\Rightarrow{ln}\left(\mathrm{1}+{s}\right)=\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} {s}^{{k}} }{{k}} \\ $$$$\Rightarrow\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} }{{k}}={ln}\left(\mathrm{2}\right)−\underset{{k}={n}+\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} }{{k}} \\ $$$$\Rightarrow{A}_{{n}} =\left(−\mathrm{1}\right)^{{n}} \underset{{k}={n}+\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} }{{k}} \\ $$$$\Rightarrow\underset{{n}\rightarrow\infty} {{lim}A}_{{n}} =\mathrm{0} \\ $$
Answered by Mathspace last updated on 25/Jun/22
![A_n =∫_R^+ (x^n /(1+x))χ_([0,1]) (x)dx =∫_R^+ f_n (x)dx f_m converge simplement vers0 car x ∈[0,1] and f_n est dominee par g(x)=(1/(1+x)) ⇒lim A_n =∫_R^+ lim f_n =0](https://www.tinkutara.com/question/Q172330.png)
$${A}_{{n}} =\int_{{R}^{+} } \:\:\:\frac{{x}^{{n}} }{\mathrm{1}+{x}}\chi_{\left[\mathrm{0},\mathrm{1}\right]} \:\left({x}\right){dx} \\ $$$$=\int_{{R}^{+} } \:\:{f}_{{n}} \left({x}\right){dx} \\ $$$${f}_{{m}} {converge}\:{simplement}\:{vers}\mathrm{0} \\ $$$${car}\:{x}\:\in\left[\mathrm{0},\mathrm{1}\right]\:\:{and}\:{f}_{{n}} {est}\:{dominee}\:{par} \\ $$$${g}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{1}+{x}}\:\Rightarrow{lim}\:{A}_{{n}} =\int_{{R}^{+} } \:{lim}\:{f}_{{n}} =\mathrm{0} \\ $$
Answered by puissant last updated on 25/Jun/22
$$\:\mathrm{0}\leqslant{x}\leqslant\mathrm{1}\:\:\:\:\:\:\:\:\:\Rightarrow\:\:\:\:\:\:\:\mathrm{1}\leqslant\mathrm{1}+{x}\leqslant\mathrm{2} \\ $$$$\:{on}\:{a}\::\:\:\frac{{x}^{{n}} }{\mathrm{2}}\leqslant\:\frac{{x}^{{n}} }{\mathrm{1}+{x}}\leqslant{x}^{{n}} \\ $$$$\Rightarrow\:\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{n}} {dx}\:\leqslant\:{I}_{{n}} \leqslant\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{n}} {dx} \\ $$$$\Rightarrow\:\frac{\mathrm{1}}{\mathrm{2}\left({n}+\mathrm{1}\right)}\:\leqslant\:{I}_{{n}} \leqslant\:\frac{\mathrm{1}}{{n}+\mathrm{1}} \\ $$$${donc}\:\underset{{n}\rightarrow+\infty} {\mathrm{lim}}{I}_{{n}} =\mathrm{0}\: \\ $$$${d}'{apres}\:{le}\:{theoreme}\:{des}\:{gendarmes}.. \\ $$