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Question Number 202865 by Mathspace last updated on 04/Jan/24

$$findtheseauence{u}_{n}wichverify$$$$u_0=1and{u}_n+u_{n+1}=\frac{{(-1)}^n}{n}$$$$forn⩾1$$

Commented by mr W last updated on 05/Jan/24

$$youmean{u}_1=1?$$

Answered by witcher3 last updated on 04/Jan/24

$$\mathrm{not}\mathrm{defined}$$$$\mathrm{may}\mathrm{bee}{\mathrm{u}}_{\mathrm{n}}+{\mathrm{u}}_{\mathrm{n}+1}=\frac{{(-1)}^{\mathrm{n}+1}}{\mathrm{n}+1}$$$$$$

Commented by Mathspace last updated on 05/Jan/24

$$nosirthissequenceisdefined...$$

Answered by Mathspace last updated on 05/Jan/24

$$take{u}_1=1$$

Answered by witcher3 last updated on 05/Jan/24

$${(-1)}^{\mathrm{n}}{\mathrm{u}}_{\mathrm{n}}-{(-1)}^{\mathrm{n}+1}{\mathrm{u}}_{\mathrm{n}+1}=\frac{1}{\mathrm{n}}$$$${(-1)}^{\mathrm{n}}{\mathrm{u}}_{\mathrm{n}}={\mathrm{v}}_{\mathrm{n}}$$$$\Rightarrow {\mathrm{v}}_{\mathrm{n}}-{\mathrm{v}}_{\mathrm{n}+1}=\frac{1}{\mathrm{n}};\mathrm{\forall }\mathrm{n}⩾1$$$$\underset{\mathrm{k}=1}{\sum ^{\mathrm{n}-1}}{\mathrm{v}}_{\mathrm{k}}-{\mathrm{v}}_{\mathrm{k}+1}=\underset{\mathrm{k}=1}{\sum ^{\mathrm{n}-1}}\frac{1}{\mathrm{k}};\mathrm{n}⩾2$$$$\Rightarrow {\mathrm{v}}_1-{\mathrm{v}}_{\mathrm{n}}={\mathrm{H}}_{\mathrm{n}-1}$$$${\mathrm{v}}_{\mathrm{n}}={\mathrm{v}}_1-{\mathrm{H}}_{\mathrm{n}-1};{\mathrm{v}}_1=-{\mathrm{u}}_1=-1$$$${(-1)}^{\mathrm{n}}(-1-{\mathrm{H}}_{\mathrm{n}-1})={\mathrm{u}}_{\mathrm{n}};\mathrm{\forall }\mathrm{n}⩾2$$$${\mathrm{u}}_1=1$$$${\mathrm{u}}_{\mathrm{n}}=\{\begin{array}{l}{(-1)}^{\mathrm{n}}(-1-{\mathrm{H}}_{\mathrm{n}-1});\mathrm{n}⩾2\\ 1,\mathrm{n}=1\end{array}$$

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