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Question Number 38804 by maxmathsup by imad last updated on 30/Jun/18

let A_n = ∫_0 ^n (((−1)^x )/(2[x] +1))dx 1) calculate A_n 2) find lim_(n→+∞) A_n

$${let}\:\:{A}_{{n}} =\:\int_{\mathrm{0}} ^{{n}} \:\:\:\frac{\left(−\mathrm{1}\right)^{{x}} }{\mathrm{2}\left[{x}\right]\:+\mathrm{1}}{dx} \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{A}_{{n}} \\ $$$$\left.\mathrm{2}\right)\:{find}\:{lim}_{{n}\rightarrow+\infty} \:{A}_{{n}} \\ $$

Commented by math khazana by abdo last updated on 30/Jun/18

A_n = ∫_0 ^n (((−1)^([x]) )/(2[x]+1))dx

$${A}_{{n}} =\:\int_{\mathrm{0}} ^{{n}} \:\:\:\frac{\left(−\mathrm{1}\right)^{\left[{x}\right]} }{\mathrm{2}\left[{x}\right]+\mathrm{1}}{dx} \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 30/Jun/18

∫_0 ^1 (((−1)^0 )/(2×0+2))dx+∫_1 ^2 (((−1)^1 )/(2×1+1))+∫_2 ^3 (((−1)^2 )/(2×2))+...+

$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\left(−\mathrm{1}\right)^{\mathrm{0}} }{\mathrm{2}×\mathrm{0}+\mathrm{2}}{dx}+\int_{\mathrm{1}} ^{\mathrm{2}} \frac{\left(−\mathrm{1}\right)^{\mathrm{1}} }{\mathrm{2}×\mathrm{1}+\mathrm{1}}+\int_{\mathrm{2}} ^{\mathrm{3}} \frac{\left(−\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{2}×\mathrm{2}}+...+ \\ $$

Commented by math khazana by abdo last updated on 01/Jul/18

A_n =Σ_(k=0) ^(n−1) ∫_k ^(k+1) (((−1)^k )/(2k+1))dx =Σ_(k=0) ^(n−1) (((−1)^k )/(2k+1)) 2)lim_(n→+∞) A_n = Σ_(k=0) ^∞ (((−1)^k )/(2k+1)) =(π/4) .

$${A}_{{n}} =\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\int_{{k}} ^{{k}+\mathrm{1}} \:\frac{\left(−\mathrm{1}\right)^{{k}} }{\mathrm{2}{k}+\mathrm{1}}{dx}\:=\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\frac{\left(−\mathrm{1}\right)^{{k}} }{\mathrm{2}{k}+\mathrm{1}} \\ $$$$\left.\mathrm{2}\right){lim}_{{n}\rightarrow+\infty} \:{A}_{{n}} =\:\sum_{{k}=\mathrm{0}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{k}} }{\mathrm{2}{k}+\mathrm{1}}\:=\frac{\pi}{\mathrm{4}}\:. \\ $$$$ \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 30/Jun/18

∫_0 ^1 (((−1)^0 )/(2×0+1))dx+∫_1 ^2 (((−1)^1 )/(2×1+1))dx+∫_2 ^3 (((−1)^2 )/(2×2+1))dx+... +∫_(n−1) ^n (((−1)^n )/(2(n−1)+1))dx =(1/1)−(1/3)+(1/5)+...+(((−1)^n )/(2n−1)) 2)we know tan^(−1) x=x−(x^3 /3)+(x^5 /5)−(x^7 /7)+... so when n→∞ A_n →tan^(−1) (1) so ans is n→∞ A_n →(Π/4)

$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\left(−\mathrm{1}\right)^{\mathrm{0}} }{\mathrm{2}×\mathrm{0}+\mathrm{1}}{dx}+\int_{\mathrm{1}} ^{\mathrm{2}} \frac{\left(−\mathrm{1}\right)^{\mathrm{1}} }{\mathrm{2}×\mathrm{1}+\mathrm{1}}{dx}+\int_{\mathrm{2}} ^{\mathrm{3}} \frac{\left(−\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{2}×\mathrm{2}+\mathrm{1}}{dx}+... \\ $$$$\:+\int_{{n}−\mathrm{1}} ^{{n}} \frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}\left({n}−\mathrm{1}\right)+\mathrm{1}}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{5}}+...+\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}{n}−\mathrm{1}} \\ $$$$\left.\mathrm{2}\right){we}\:{know}\:{tan}^{−\mathrm{1}} {x}={x}−\frac{{x}^{\mathrm{3}} }{\mathrm{3}}+\frac{{x}^{\mathrm{5}} }{\mathrm{5}}−\frac{{x}^{\mathrm{7}} }{\mathrm{7}}+... \\ $$$${so}\:{when}\:{n}\rightarrow\infty\:\:{A}_{{n}} \rightarrow{tan}^{−\mathrm{1}} \left(\mathrm{1}\right) \\ $$$${so}\:{ans}\:{is}\: \\ $$$${n}\rightarrow\infty\:{A}_{{n}} \rightarrow\frac{\Pi}{\mathrm{4}} \\ $$$$ \\ $$

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