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Question Number 57847 by Abdo msup. last updated on 13/Apr/19

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

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

Commented by maxmathsup by imad last updated on 14/Apr/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) (((−1)^k )/k^2 ) ⇒ −(u_1 +u_2 )+u_2 +u_3 −(u_3 +u_4 ) +....(−1)^(n−1) (u_(n−1) +u_n ) =Σ_(k=1) ^(n−1) (((−1)^k )/k^2 ) ⇒ −u_1 +(−1)^(n−1) u_n =Σ_(k=1) ^(n−1) (((−1)^k )/k^2 ) ⇒ (−1)^(n−1) u_n =Σ_(k=1) ^(n−1) (((−1)^k )/k^2 ) +u_1 ⇒ u_n =Σ_(k=1) ^(n−1) (((−1)^(k−n+1) )/k^2 ) +(−1)^(n−1) u_1 with n≥2 2)the sequence (−1)^n u_1 is divergent so (u_n ) diverges but we can see u_(2n) =−Σ_(k=1) ^(2n−1) (((−1)^k )/k^2 ) −u_1 →−Σ_(k=1) ^∞ (((−1)^k )/k^2 ) −u_1 ⇒(u_(2n) ) converges also u_(2n+1) =Σ_(k=1) ^(2n) (((−1)^k )/k^2 ) +u_1 →Σ_(k=1) ^∞ (((−1)^k )/k^2 ) +u_1 ⇒u_(2n+1) converges

$$\left.\mathrm{1}\right)\:{we}\:{have}\:{u}_{{n}} +{u}_{{n}+\mathrm{1}} =\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\:\:\Rightarrow\sum_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} \:\left(−\mathrm{1}\right)^{{k}} \left({u}_{{k}} \:+{u}_{{k}+\mathrm{1}} \right)=\sum_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} \:\frac{\left(−\mathrm{1}\right)^{{k}} }{{k}^{\mathrm{2}} }\:\Rightarrow \\ $$$$−\left({u}_{\mathrm{1}} \:+{u}_{\mathrm{2}} \right)+{u}_{\mathrm{2}} \:+{u}_{\mathrm{3}} −\left({u}_{\mathrm{3}} \:+{u}_{\mathrm{4}} \right)\:+....\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \left({u}_{{n}−\mathrm{1}} \:+{u}_{{n}} \right)\:=\sum_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} \:\frac{\left(−\mathrm{1}\right)^{{k}} }{{k}^{\mathrm{2}} }\:\Rightarrow \\ $$$$−{u}_{\mathrm{1}} \:+\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \:{u}_{{n}} =\sum_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} \:\frac{\left(−\mathrm{1}\right)^{{k}} }{{k}^{\mathrm{2}} }\:\Rightarrow\:\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} {u}_{{n}} =\sum_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} \:\frac{\left(−\mathrm{1}\right)^{{k}} }{{k}^{\mathrm{2}} }\:+{u}_{\mathrm{1}} \:\Rightarrow \\ $$$${u}_{{n}} =\sum_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} \:\:\frac{\left(−\mathrm{1}\right)^{{k}−{n}+\mathrm{1}} }{{k}^{\mathrm{2}} }\:+\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} {u}_{\mathrm{1}} \:\:\:\:{with}\:{n}\geqslant\mathrm{2} \\ $$$$\left.\mathrm{2}\right){the}\:{sequence}\:\left(−\mathrm{1}\right)^{{n}} {u}_{\mathrm{1}} \:\:{is}\:{divergent}\:\:{so}\:\left({u}_{{n}} \right)\:{diverges}\:\:{but}\:{we}\:{can}\:{see}\: \\ $$$${u}_{\mathrm{2}{n}} =−\sum_{{k}=\mathrm{1}} ^{\mathrm{2}{n}−\mathrm{1}} \:\frac{\left(−\mathrm{1}\right)^{{k}} }{{k}^{\mathrm{2}} }\:−{u}_{\mathrm{1}} \:\:\rightarrow−\sum_{{k}=\mathrm{1}} ^{\infty} \:\:\:\frac{\left(−\mathrm{1}\right)^{{k}} }{{k}^{\mathrm{2}} }\:−{u}_{\mathrm{1}} \:\Rightarrow\left({u}_{\mathrm{2}{n}} \right)\:{converges}\:{also} \\ $$$${u}_{\mathrm{2}{n}+\mathrm{1}} =\sum_{{k}=\mathrm{1}} ^{\mathrm{2}{n}} \:\:\frac{\left(−\mathrm{1}\right)^{{k}} }{{k}^{\mathrm{2}} }\:+{u}_{\mathrm{1}} \rightarrow\sum_{{k}=\mathrm{1}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{k}} }{{k}^{\mathrm{2}} }\:+{u}_{\mathrm{1}} \:\Rightarrow{u}_{\mathrm{2}{n}+\mathrm{1}} \:{converges} \\ $$

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