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Question Number 65489 by mathmax by abdo last updated on 30/Jul/19

U_n is a sequence wich verify U_n +U_(n+1) =(1/n^2 ) 1) find U_n interms of n 2) calculate lim_(n→+∞) U_n

$${U}_{{n}} \:{is}\:{a}\:{sequence}\:{wich}\:{verify}\:\:{U}_{{n}} \:+{U}_{{n}+\mathrm{1}} =\frac{\mathrm{1}}{{n}^{\mathrm{2}} } \\ $$$$\left.\mathrm{1}\right)\:{find}\:{U}_{{n}} \:{interms}\:{of}\:{n} \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:{lim}_{{n}\rightarrow+\infty} \:{U}_{{n}} \\ $$

Commented by mathmax by abdo last updated on 03/Aug/19

1) we have u_n +u_(n+1) =(1/n^2 ) ⇒Σ_(k=1) ^(n−1) (−1)^k (u_k +u_(k+1) )=Σ_(k=1) ^n (((−1)^k )/k^2 ) ⇒−u_1 −u_2 +u_2 +u_3 −....+(−1)^(n−2) (u_(n−2) +u_(n−1) )+ (−1)^(n−1) (u_(n−1) +u_n )=Σ_(k=1) ^n (((−1)^k )/k^2 ) ⇒ −u_1 +(−1)^(n−1) u_n =Σ_(k=1) ^n (((−1)^k )/k^2 ) ⇒ (−1)^(n−1) u_n =Σ_(k=1) ^n (((−1)^k )/k^2 ) +u_1 ⇒u_n =(−1)^(n−1) Σ_(k=1) ^n (((−1)^k )/k^2 ) +(−1)^(n−1) u_1 2) u_n is not convergent but we see that u_(2n) =−Σ_(k=1) ^(2n) (((−1)^k )/k^2 ) −u_1 →−Σ_(k=1) ^∞ (((−1)^k )/k^2 )−u_1 u_(2n+1) =Σ_(k=1) ^(2n+1) (((−1)^k )/k^2 ) +u_1 →Σ_(k=1) ^∞ (((−1)^k )/k^2 ) +u_1

$$\left.\mathrm{1}\right)\:{we}\:{have}\:{u}_{{n}} +{u}_{{n}+\mathrm{1}} =\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\:\Rightarrow\sum_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} \left(−\mathrm{1}\right)^{{k}} \left({u}_{{k}} +{u}_{{k}+\mathrm{1}} \right)=\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\left(−\mathrm{1}\right)^{{k}} }{{k}^{\mathrm{2}} } \\ $$$$\Rightarrow−{u}_{\mathrm{1}} −{u}_{\mathrm{2}} \:+{u}_{\mathrm{2}} \:+{u}_{\mathrm{3}} \:−....+\left(−\mathrm{1}\right)^{{n}−\mathrm{2}} \left({u}_{{n}−\mathrm{2}} \:+{u}_{{n}−\mathrm{1}} \right)+ \\ $$$$\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \left({u}_{{n}−\mathrm{1}} \:+{u}_{{n}} \right)=\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\left(−\mathrm{1}\right)^{{k}} }{{k}^{\mathrm{2}} }\:\Rightarrow \\ $$$$−{u}_{\mathrm{1}} +\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} {u}_{{n}} =\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\left(−\mathrm{1}\right)^{{k}} }{{k}^{\mathrm{2}} }\:\Rightarrow \\ $$$$\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} {u}_{{n}} =\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\left(−\mathrm{1}\right)^{{k}} }{{k}^{\mathrm{2}} }\:+{u}_{\mathrm{1}} \:\Rightarrow{u}_{{n}} =\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\left(−\mathrm{1}\right)^{{k}} }{{k}^{\mathrm{2}} }\:+\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} {u}_{\mathrm{1}} \\ $$$$\left.\mathrm{2}\right)\:{u}_{{n}} \:{is}\:{not}\:{convergent}\:\:{but}\:{we}\:{see}\:{that} \\ $$$${u}_{\mathrm{2}{n}} =−\sum_{{k}=\mathrm{1}} ^{\mathrm{2}{n}} \:\frac{\left(−\mathrm{1}\right)^{{k}} }{{k}^{\mathrm{2}} }\:−{u}_{\mathrm{1}} \rightarrow−\sum_{{k}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{k}} }{{k}^{\mathrm{2}} }−{u}_{\mathrm{1}} \\ $$$${u}_{\mathrm{2}{n}+\mathrm{1}} =\sum_{{k}=\mathrm{1}} ^{\mathrm{2}{n}+\mathrm{1}} \:\frac{\left(−\mathrm{1}\right)^{{k}} }{{k}^{\mathrm{2}} }\:+{u}_{\mathrm{1}} \rightarrow\sum_{{k}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{k}} }{{k}^{\mathrm{2}} }\:+{u}_{\mathrm{1}} \\ $$

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