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Question Number 45599 by maxmathsup by imad last updated on 14/Oct/18
![1) calculate A_n = ∫_0 ^n (((−1)^([x]) )/(2x+1−[x]))dx 2) find lim_(n→+∞) A_n 3) study the serie Σ A_n](Q45599.png)
$$\left.\mathrm{1}\right)\:{calculate}\:{A}_{{n}} =\:\int_{\mathrm{0}} ^{{n}} \:\:\:\:\frac{\left(−\mathrm{1}\right)^{\left[{x}\right]} }{\mathrm{2}{x}+\mathrm{1}−\left[{x}\right]}{dx} \\ $$$$\left.\mathrm{2}\right)\:{find}\:{lim}_{{n}\rightarrow+\infty} {A}_{{n}} \\ $$$$\left.\mathrm{3}\right)\:{study}\:{the}\:{serie}\:\:\Sigma\:{A}_{{n}} \\ $$
Commented by maxmathsup by imad last updated on 21/Oct/18
![1) we have A_n =Σ_(k=0) ^(n−1) ∫_k ^(k+1) (((−1)^k )/(2x+1−k))dx =Σ_(k=0) ^(n−1) (−1)^k ∫_k ^(k+1) (dx/(2x+1−k)) =Σ_(k=0) ^(n−1) (((−1)^k )/2)[ln∣2x+1−k∣]_k ^(k+1) =Σ_(k=0) ^(n−1) (((−1^k )/2){ln(k+3)−ln(k+1)} =Σ_(k=0) ^(n−1) (((−1)^k )/2)ln(((k+3)/(k+1)))⇒ A_n =(1/2){ln(3)−ln(2) +ln((5/3))−ln((6/4))+...(−1)^(n−1) ln(((n+2)/n))}](Q46115.png)
$$\left.\mathrm{1}\right)\:{we}\:{have}\:{A}_{{n}} =\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\:\:\int_{{k}} ^{{k}+\mathrm{1}} \:\:\:\frac{\left(−\mathrm{1}\right)^{{k}} }{\mathrm{2}{x}+\mathrm{1}−{k}}{dx} \\ $$$$=\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\left(−\mathrm{1}\right)^{{k}} \:\int_{{k}} ^{{k}+\mathrm{1}} \:\:\:\frac{{dx}}{\mathrm{2}{x}+\mathrm{1}−{k}}\:=\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\frac{\left(−\mathrm{1}\right)^{{k}} }{\mathrm{2}}\left[{ln}\mid\mathrm{2}{x}+\mathrm{1}−{k}\mid\right]_{{k}} ^{{k}+\mathrm{1}} \\ $$$$=\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\:\:\frac{\left(−\mathrm{1}^{{k}} \right.}{\mathrm{2}}\left\{{ln}\left({k}+\mathrm{3}\right)−{ln}\left({k}+\mathrm{1}\right)\right\} \\ $$$$=\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\:\frac{\left(−\mathrm{1}\right)^{{k}} }{\mathrm{2}}{ln}\left(\frac{{k}+\mathrm{3}}{{k}+\mathrm{1}}\right)\Rightarrow \\ $$$${A}_{{n}} =\frac{\mathrm{1}}{\mathrm{2}}\left\{{ln}\left(\mathrm{3}\right)−{ln}\left(\mathrm{2}\right)\:+{ln}\left(\frac{\mathrm{5}}{\mathrm{3}}\right)−{ln}\left(\frac{\mathrm{6}}{\mathrm{4}}\right)+...\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} {ln}\left(\frac{{n}+\mathrm{2}}{{n}}\right)\right\} \\ $$