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Question Number 61536 by maxmathsup by imad last updated on 04/Jun/19

1)let U_n =Σ_(k=0) ^n (−1)^k  =1−1+1−1+...(n+1 terms)  is lim_(n→+∞) U_n exist ?  find U_n  by using integr part[..]  2) let V_n = Σ_(k=1) ^n k(−1)^k   = −1+2 −3+4+.....(nterms)  is lim_(n→+∞) V_n  exist   find V_n by using integr part[..]

$$\left.\mathrm{1}\right){let}\:{U}_{{n}} =\sum_{{k}=\mathrm{0}} ^{{n}} \left(−\mathrm{1}\right)^{{k}} \:=\mathrm{1}−\mathrm{1}+\mathrm{1}−\mathrm{1}+...\left({n}+\mathrm{1}\:{terms}\right) \\ $$$${is}\:{lim}_{{n}\rightarrow+\infty} {U}_{{n}} {exist}\:?\:\:{find}\:{U}_{{n}} \:{by}\:{using}\:{integr}\:{part}\left[..\right] \\ $$$$\left.\mathrm{2}\right)\:{let}\:{V}_{{n}} =\:\sum_{{k}=\mathrm{1}} ^{{n}} {k}\left(−\mathrm{1}\right)^{{k}} \:\:=\:−\mathrm{1}+\mathrm{2}\:−\mathrm{3}+\mathrm{4}+.....\left({nterms}\right) \\ $$$${is}\:{lim}_{{n}\rightarrow+\infty} {V}_{{n}} \:{exist}\: \\ $$$${find}\:{V}_{{n}} {by}\:{using}\:{integr}\:{part}\left[..\right] \\ $$

Commented by maxmathsup by imad last updated on 04/Jun/19

1) we have for x≠1   Σ_(k=0) ^n  x^k  =((1−x^(n+1) )/(1−x)) ⇒Σ_(k=0) ^n (−1)^k  =((1−(−1)^(n+1) )/2)  ⇒ U_n =(1/2) +(((−1)^n )/2)     the sequence (−1)^n  is not convergent  so U_n dont converges  but we have  lim_(n→+∞)    U_(2n) =1   and lim_(n→+∞)   U_(2n+1) =0  2) we have V_n =w(−1)  with  w(x) =Σ_(k=1) ^n  k x^k  =Σ_(k=0) ^n  kx^k   we have Σ_(k=0) ^n  x^k  =((x^(n+1) −1)/(x−1)) ⇒Σ_(k=1) ^n  kx^(k−1)  =(((n+1)x^n (x−1)−(x^(n+1) −1)×1)/((x−1)^2 ))  =(((n+1)x^(n+1) −(n+1)x^n −x^(n+1)  +1)/((x−1)^2 )) =((nx^(n+1) −(n+1)x^n  +1)/((x−1)^2 ))  ⇒  V_n =((n(−1)^(n+1) −(n+1)(−1)^n  +1)/4) =((−n(−1)^n −(n+1)(−1)^n  +1)/4)  =((1−(2n+1)(−1)^n )/4)   its clear that V_n  is not convergent but we see  V_(2n) =((1−(2n+1))/4) →−∞    and  V_(2n+1) =((1+2n+1)/4) =((n+1)/2) →+∞   (n→+∞)

$$\left.\mathrm{1}\right)\:{we}\:{have}\:{for}\:{x}\neq\mathrm{1}\:\:\:\sum_{{k}=\mathrm{0}} ^{{n}} \:{x}^{{k}} \:=\frac{\mathrm{1}−{x}^{{n}+\mathrm{1}} }{\mathrm{1}−{x}}\:\Rightarrow\sum_{{k}=\mathrm{0}} ^{{n}} \left(−\mathrm{1}\right)^{{k}} \:=\frac{\mathrm{1}−\left(−\mathrm{1}\right)^{\boldsymbol{{n}}+\mathrm{1}} }{\mathrm{2}} \\ $$$$\Rightarrow\:{U}_{{n}} =\frac{\mathrm{1}}{\mathrm{2}}\:+\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}}\:\:\:\:\:{the}\:{sequence}\:\left(−\mathrm{1}\right)^{{n}} \:{is}\:{not}\:{convergent}\:\:{so}\:{U}_{{n}} {dont}\:{converges} \\ $$$${but}\:{we}\:{have}\:\:{lim}_{{n}\rightarrow+\infty} \:\:\:{U}_{\mathrm{2}{n}} =\mathrm{1}\:\:\:{and}\:{lim}_{{n}\rightarrow+\infty} \:\:{U}_{\mathrm{2}{n}+\mathrm{1}} =\mathrm{0} \\ $$$$\left.\mathrm{2}\right)\:{we}\:{have}\:{V}_{{n}} ={w}\left(−\mathrm{1}\right)\:\:{with}\:\:{w}\left({x}\right)\:=\sum_{{k}=\mathrm{1}} ^{{n}} \:{k}\:{x}^{{k}} \:=\sum_{{k}=\mathrm{0}} ^{{n}} \:{kx}^{{k}} \\ $$$${we}\:{have}\:\sum_{{k}=\mathrm{0}} ^{{n}} \:{x}^{{k}} \:=\frac{{x}^{{n}+\mathrm{1}} −\mathrm{1}}{{x}−\mathrm{1}}\:\Rightarrow\sum_{{k}=\mathrm{1}} ^{{n}} \:{kx}^{{k}−\mathrm{1}} \:=\frac{\left({n}+\mathrm{1}\right){x}^{{n}} \left({x}−\mathrm{1}\right)−\left({x}^{{n}+\mathrm{1}} −\mathrm{1}\right)×\mathrm{1}}{\left({x}−\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$=\frac{\left({n}+\mathrm{1}\right){x}^{{n}+\mathrm{1}} −\left({n}+\mathrm{1}\right){x}^{{n}} −{x}^{{n}+\mathrm{1}} \:+\mathrm{1}}{\left({x}−\mathrm{1}\right)^{\mathrm{2}} }\:=\frac{{nx}^{{n}+\mathrm{1}} −\left({n}+\mathrm{1}\right){x}^{{n}} \:+\mathrm{1}}{\left({x}−\mathrm{1}\right)^{\mathrm{2}} }\:\:\Rightarrow \\ $$$${V}_{{n}} =\frac{{n}\left(−\mathrm{1}\right)^{\boldsymbol{{n}}+\mathrm{1}} −\left(\boldsymbol{{n}}+\mathrm{1}\right)\left(−\mathrm{1}\right)^{\boldsymbol{{n}}} \:+\mathrm{1}}{\mathrm{4}}\:=\frac{−{n}\left(−\mathrm{1}\right)^{{n}} −\left({n}+\mathrm{1}\right)\left(−\mathrm{1}\right)^{{n}} \:+\mathrm{1}}{\mathrm{4}} \\ $$$$=\frac{\mathrm{1}−\left(\mathrm{2}{n}+\mathrm{1}\right)\left(−\mathrm{1}\right)^{{n}} }{\mathrm{4}}\:\:\:{its}\:{clear}\:{that}\:\boldsymbol{{V}}_{{n}} \:{is}\:{not}\:{convergent}\:{but}\:{we}\:{see} \\ $$$${V}_{\mathrm{2}{n}} =\frac{\mathrm{1}−\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{4}}\:\rightarrow−\infty\:\:\:\:{and}\:\:{V}_{\mathrm{2}{n}+\mathrm{1}} =\frac{\mathrm{1}+\mathrm{2}{n}+\mathrm{1}}{\mathrm{4}}\:=\frac{{n}+\mathrm{1}}{\mathrm{2}}\:\rightarrow+\infty\:\:\:\left({n}\rightarrow+\infty\right) \\ $$

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