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Question Number 124065 by Bird last updated on 30/Nov/20

determine tbe sewuence u_n wich verify u_n +u_(n+1) =(((−1)^n )/( (√n))) n>0

$${determine}\:{tbe}\:{sewuence}\:{u}_{{n}} \\ $$ $${wich}\:{verify}\:{u}_{{n}} +{u}_{{n}+\mathrm{1}} =\frac{\left(−\mathrm{1}\right)^{{n}} }{\:\sqrt{{n}}} \\ $$ $${n}>\mathrm{0} \\ $$

Answered by mindispower last updated on 30/Nov/20

(−1)^n u_n +(−1)^n u_(n+1) =(1/( (√n))) Σ_(k=1) ^n (−1)^k u_k +(−1)^k u_(k+1) =Σ_(k=1) ^n (1/( (√k))) (−1)^n u_(n+1) −u_1 =Σ_(k=1) ^n (1/( (√k))) u_n =(−1)^(n−1) u_1 +(−1)^(n−1) Σ_(k≤n−1) (1/( (√k))),∀n≥2

$$\left(−\mathrm{1}\right)^{{n}} {u}_{{n}} +\left(−\mathrm{1}\right)^{{n}} {u}_{{n}+\mathrm{1}} =\frac{\mathrm{1}}{\:\sqrt{{n}}} \\ $$ $$\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left(−\mathrm{1}\right)^{{k}} {u}_{{k}} +\left(−\mathrm{1}\right)^{{k}} {u}_{{k}+\mathrm{1}} =\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{\:\sqrt{{k}}} \\ $$ $$\left(−\mathrm{1}\right)^{{n}} {u}_{{n}+\mathrm{1}} −{u}_{\mathrm{1}} =\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{\:\sqrt{{k}}} \\ $$ $${u}_{{n}} =\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} {u}_{\mathrm{1}} +\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \underset{{k}\leqslant{n}−\mathrm{1}} {\sum}\frac{\mathrm{1}}{\:\sqrt{{k}}},\forall{n}\geqslant\mathrm{2} \\ $$ $$ \\ $$

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