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1-let-U-n-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-k-1-n-k-1-k-1-2-3-4-nterms-is-lim-n-V-n-exist

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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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