Question Number 39703 by math khazana by abdo last updated on 10/Jul/18
![let A_n = ∫_0 ^n (((−1)^([x]) )/(x+2−[x]))dx 1) calculate A_n and lim_(n→+∞) A_n 2) let S_n =Σ_(n=0) ^n A_n dtudy the convergence of S_n 4) let W_n = Σ_(n=1) ^n (1/A_n ) study the convergence of W_n](https://www.tinkutara.com/question/Q39703.png)
$${let}\:\:{A}_{{n}} =\:\int_{\mathrm{0}} ^{{n}} \:\:\:\frac{\left(−\mathrm{1}\right)^{\left[{x}\right]} }{{x}+\mathrm{2}−\left[{x}\right]}{dx} \\ $$$$\left.\mathrm{1}\right)\:\:{calculate}\:{A}_{{n}} \:\:\:{and}\:{lim}_{{n}\rightarrow+\infty} \:\:{A}_{{n}} \: \\ $$$$\left.\mathrm{2}\right)\:{let}\:{S}_{{n}} \:=\sum_{{n}=\mathrm{0}} ^{{n}} \:\:{A}_{{n}} \:\:\:{dtudy}\:{the}\:{convergence}\:{of}\:{S}_{{n}} \\ $$$$\left.\mathrm{4}\right)\:{let}\:\:{W}_{{n}} =\:\sum_{{n}=\mathrm{1}} ^{{n}} \:\:\frac{\mathrm{1}}{{A}_{{n}} } \\ $$$${study}\:{the}\:{convergence}\:{of}\:{W}_{{n}} \\ $$
Commented by math khazana by abdo last updated on 13/Jul/18
![1) we have A_n =Σ_(k=0) ^n ∫_k ^(k+1) (((−1)^k )/(x+2−k))dx =Σ_(k=0) ^n (−1)^k [ln∣x+2−k∣]_k ^(k+1) = Σ_(k=0) ^n (−1)^k {ln(k+1+2−k) −ln(k+2−k) =Σ_(k=0) ^n (−1)^k {ln((3/2))}=ln((3/2))((1−(−1)^(n+1) )/2) ⇒ A_n =(1/2)ln((3/2))(1−(−1)^(n+1) ) but the sequence (−1)^n is not convegent so A_n is not convergent! 2) S_n =Σ_(n=0) ^∞ A_n =Σ_(n=0) ^∞ (1/2)ln((3/2)){1+(−1)^n } S_n also diverges 4) we have A_n =0 if n even and A_n =ln((3/2)) if n is odd so W_n is not defined!](https://www.tinkutara.com/question/Q39931.png)
$$\left.\mathrm{1}\right)\:{we}\:{have}\:{A}_{{n}} =\sum_{{k}=\mathrm{0}} ^{{n}} \:\:\int_{{k}} ^{{k}+\mathrm{1}} \:\:\:\:\frac{\left(−\mathrm{1}\right)^{{k}} }{{x}+\mathrm{2}−{k}}{dx} \\ $$$$=\sum_{{k}=\mathrm{0}} ^{{n}} \:\left(−\mathrm{1}\right)^{{k}} \left[{ln}\mid{x}+\mathrm{2}−{k}\mid\right]_{{k}} ^{{k}+\mathrm{1}} \\ $$$$=\:\sum_{{k}=\mathrm{0}} ^{{n}} \:\left(−\mathrm{1}\right)^{{k}} \left\{{ln}\left({k}+\mathrm{1}+\mathrm{2}−{k}\right)\:−{ln}\left({k}+\mathrm{2}−{k}\right)\right. \\ $$$$=\sum_{{k}=\mathrm{0}} ^{{n}} \left(−\mathrm{1}\right)^{{k}} \left\{{ln}\left(\frac{\mathrm{3}}{\mathrm{2}}\right)\right\}={ln}\left(\frac{\mathrm{3}}{\mathrm{2}}\right)\frac{\mathrm{1}−\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} }{\mathrm{2}} \\ $$$$\Rightarrow\:{A}_{{n}} =\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\frac{\mathrm{3}}{\mathrm{2}}\right)\left(\mathrm{1}−\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} \right)\:{but}\:{the}\:{sequence} \\ $$$$\left(−\mathrm{1}\right)^{{n}} \:{is}\:{not}\:{convegent}\:{so}\:\:{A}_{{n}} {is}\:{not}\:{convergent}! \\ $$$$\left.\mathrm{2}\right)\:{S}_{{n}} \:=\sum_{{n}=\mathrm{0}} ^{\infty} \:{A}_{{n}} =\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\frac{\mathrm{3}}{\mathrm{2}}\right)\left\{\mathrm{1}+\left(−\mathrm{1}\right)^{{n}} \right\} \\ $$$${S}_{{n}} \:{also}\:{diverges} \\ $$$$\left.\mathrm{4}\right)\:{we}\:{have}\:{A}_{{n}} =\mathrm{0}\:{if}\:{n}\:{even}\:{and}\:{A}_{{n}} ={ln}\left(\frac{\mathrm{3}}{\mathrm{2}}\right)\:{if}\:{n} \\ $$$${is}\:{odd}\:{so}\:{W}_{{n}} \:{is}\:{not}\:{defined}! \\ $$