Question and Answers Forum

All Questions      Topic List

Relation and Functions Questions

Previous in All Question      Next in All Question      

Previous in Relation and Functions      Next in Relation and Functions      

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

find the seauence u_n wich verify u_0 =1 and u_n +u_(n+1) =(((−1)^n )/n) for n≥1

$${find}\:{the}\:{seauence}\:{u}_{{n}} {wich}\:{verify} \\ $$$${u}_{\mathrm{0}} =\mathrm{1}\:{and}\:{u}_{{n}} +{u}_{{n}+\mathrm{1}} =\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}} \\ $$$${for}\:{n}\geqslant\mathrm{1} \\ $$

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

you mean u_1 =1?

$${you}\:{mean}\:{u}_{\mathrm{1}} =\mathrm{1}? \\ $$

Answered by witcher3 last updated on 04/Jan/24

not defined may bee u_n +u_(n+1) =(((−1)^(n+1) )/(n+1))

$$\mathrm{not}\:\mathrm{defined}\: \\ $$$$\mathrm{may}\:\mathrm{bee}\:\mathrm{u}_{\mathrm{n}} +\mathrm{u}_{\mathrm{n}+\mathrm{1}} =\frac{\left(−\mathrm{1}\right)^{\mathrm{n}+\mathrm{1}} }{\mathrm{n}+\mathrm{1}} \\ $$$$ \\ $$

Commented by Mathspace last updated on 05/Jan/24

no sir this sequence isdefined...

$${no}\:{sir}\:{this}\:{sequence}\:{isdefined}... \\ $$

Answered by Mathspace last updated on 05/Jan/24

take u_1 =1

$${take}\:{u}_{\mathrm{1}} =\mathrm{1} \\ $$

Answered by witcher3 last updated on 05/Jan/24

(−1)^n u_n −(−1)^(n+1) u_(n+1) =(1/n) (−1)^n u_n =v_n ⇒v_n −v_(n+1) =(1/n);∀n≥1 Σ_(k=1) ^(n−1) v_k −v_(k+1) =Σ_(k=1) ^(n−1) (1/k);n≥2 ⇒v_1 −v_n =H_(n−1) v_n =v_1 −H_(n−1) ;v_1 =−u_1 =−1 (−1)^n (−1−H_(n−1) )=u_n ;∀n≥2 u_1 =1 u_n = { (((−1)^n (−1−H_(n−1) );n≥2)),((1,n=1)) :}

$$\left(−\mathrm{1}\right)^{\mathrm{n}} \mathrm{u}_{\mathrm{n}} −\left(−\mathrm{1}\right)^{\mathrm{n}+\mathrm{1}} \mathrm{u}_{\mathrm{n}+\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{n}} \\ $$$$\left(−\mathrm{1}\right)^{\mathrm{n}} \mathrm{u}_{\mathrm{n}} =\mathrm{v}_{\mathrm{n}} \\ $$$$\Rightarrow\mathrm{v}_{\mathrm{n}} −\mathrm{v}_{\mathrm{n}+\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{n}};\forall\mathrm{n}\geqslant\mathrm{1} \\ $$$$\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}−\mathrm{1}} {\sum}}\mathrm{v}_{\mathrm{k}} −\mathrm{v}_{\mathrm{k}+\mathrm{1}} =\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}−\mathrm{1}} {\sum}}\frac{\mathrm{1}}{\mathrm{k}};\mathrm{n}\geqslant\mathrm{2} \\ $$$$\Rightarrow\mathrm{v}_{\mathrm{1}} −\mathrm{v}_{\mathrm{n}} =\mathrm{H}_{\mathrm{n}−\mathrm{1}} \\ $$$$\mathrm{v}_{\mathrm{n}} =\mathrm{v}_{\mathrm{1}} −\mathrm{H}_{\mathrm{n}−\mathrm{1}} ;\mathrm{v}_{\mathrm{1}} =−\mathrm{u}_{\mathrm{1}} =−\mathrm{1} \\ $$$$\left(−\mathrm{1}\right)^{\mathrm{n}} \left(−\mathrm{1}−\mathrm{H}_{\mathrm{n}−\mathrm{1}} \right)=\mathrm{u}_{\mathrm{n}} ;\forall\mathrm{n}\geqslant\mathrm{2} \\ $$$$\mathrm{u}_{\mathrm{1}} =\mathrm{1} \\ $$$$\mathrm{u}_{\mathrm{n}} =\begin{cases}{\left(−\mathrm{1}\right)^{\mathrm{n}} \left(−\mathrm{1}−\mathrm{H}_{\mathrm{n}−\mathrm{1}} \right);\mathrm{n}\geqslant\mathrm{2}}\\{\mathrm{1},\mathrm{n}=\mathrm{1}}\end{cases} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com