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Question Number 1322 by 123456 last updated on 22/Jul/15

a_0 =1 a_1 =2 a_n −a_0 =a_(n−1) −a_(n−2) ,n≥2 a_(10) =? lim_(n→+∞) a_n =? lim_(n→+∞) a_(6n+r) =? r∈Z_6

$${a}_{\mathrm{0}} =\mathrm{1} \\ $$$${a}_{\mathrm{1}} =\mathrm{2} \\ $$$${a}_{{n}} −{a}_{\mathrm{0}} ={a}_{{n}−\mathrm{1}} −{a}_{{n}−\mathrm{2}} ,{n}\geqslant\mathrm{2} \\ $$$${a}_{\mathrm{10}} =? \\ $$$$\underset{{n}\rightarrow+\infty} {\mathrm{lim}}\:{a}_{{n}} =? \\ $$$$\underset{{n}\rightarrow+\infty} {\mathrm{lim}}\:{a}_{\mathrm{6}{n}+{r}} =?\:\:\:\:\:\:\:\:{r}\in\mathbb{Z}_{\mathrm{6}} \\ $$

Commented by 112358 last updated on 22/Jul/15

a_2 −1=a_1 −1⇒a_2 =2 a_3 −1=2−2⇒a_3 =1 a_4 −1=1−2⇒a_4 =0 a_5 −1=0−1⇒a_5 =0 a_6 −1=0−0⇒a_6 =1 a_7 −1=1−0⇒a_7 =2 a_8 −1=2−1⇒a_8 =2 a_9 −1=2−2⇒a_9 =1 a_(10) −1=1−2⇒a_(10) =0 Continuing with the terms for n>10 a_(11) −1=0−1⇒a_(11) =0 a_(12) −1=0−0⇒a_(12) =1 a_(13) −1=1−0⇒a_(13) =2 a_(14) −1=2−1⇒a_(14) =2 The pattern of the sequence {a_n }=1,2,2,1,0,0,1,2,2,1,0,0,1,2,2,... is oscillatory and periodic but is not convergent. Solim_(n→+∞) a_n is indeterminate . (Correct me if I′m wrong.)

$${a}_{\mathrm{2}} −\mathrm{1}={a}_{\mathrm{1}} −\mathrm{1}\Rightarrow{a}_{\mathrm{2}} =\mathrm{2} \\ $$$${a}_{\mathrm{3}} −\mathrm{1}=\mathrm{2}−\mathrm{2}\Rightarrow{a}_{\mathrm{3}} =\mathrm{1} \\ $$$${a}_{\mathrm{4}} −\mathrm{1}=\mathrm{1}−\mathrm{2}\Rightarrow{a}_{\mathrm{4}} =\mathrm{0} \\ $$$${a}_{\mathrm{5}} −\mathrm{1}=\mathrm{0}−\mathrm{1}\Rightarrow{a}_{\mathrm{5}} =\mathrm{0} \\ $$$${a}_{\mathrm{6}} −\mathrm{1}=\mathrm{0}−\mathrm{0}\Rightarrow{a}_{\mathrm{6}} =\mathrm{1} \\ $$$${a}_{\mathrm{7}} −\mathrm{1}=\mathrm{1}−\mathrm{0}\Rightarrow{a}_{\mathrm{7}} =\mathrm{2} \\ $$$${a}_{\mathrm{8}} −\mathrm{1}=\mathrm{2}−\mathrm{1}\Rightarrow{a}_{\mathrm{8}} =\mathrm{2} \\ $$$${a}_{\mathrm{9}} −\mathrm{1}=\mathrm{2}−\mathrm{2}\Rightarrow{a}_{\mathrm{9}} =\mathrm{1} \\ $$$${a}_{\mathrm{10}} −\mathrm{1}=\mathrm{1}−\mathrm{2}\Rightarrow{a}_{\mathrm{10}} =\mathrm{0} \\ $$$${Continuing}\:{with}\:{the}\:{terms}\:{for}\:{n}>\mathrm{10} \\ $$$${a}_{\mathrm{11}} −\mathrm{1}=\mathrm{0}−\mathrm{1}\Rightarrow{a}_{\mathrm{11}} =\mathrm{0} \\ $$$${a}_{\mathrm{12}} −\mathrm{1}=\mathrm{0}−\mathrm{0}\Rightarrow{a}_{\mathrm{12}} =\mathrm{1} \\ $$$${a}_{\mathrm{13}} −\mathrm{1}=\mathrm{1}−\mathrm{0}\Rightarrow{a}_{\mathrm{13}} =\mathrm{2} \\ $$$${a}_{\mathrm{14}} −\mathrm{1}=\mathrm{2}−\mathrm{1}\Rightarrow{a}_{\mathrm{14}} =\mathrm{2} \\ $$$${The}\:{pattern}\:{of}\:{the}\:{sequence} \\ $$$$\left\{{a}_{{n}} \right\}=\mathrm{1},\mathrm{2},\mathrm{2},\mathrm{1},\mathrm{0},\mathrm{0},\mathrm{1},\mathrm{2},\mathrm{2},\mathrm{1},\mathrm{0},\mathrm{0},\mathrm{1},\mathrm{2},\mathrm{2},... \\ $$$${is}\:{oscillatory}\:{and}\:{periodic}\:{but}\:{is}\:{not}\:{convergent}. \\ $$$${So}\underset{{n}\rightarrow+\infty} {\mathrm{lim}}{a}_{{n}} \:{is}\:{indeterminate}\:. \\ $$$$\left({Correct}\:{me}\:{if}\:{I}'{m}\:{wrong}.\right) \\ $$$$ \\ $$

Commented by 112358 last updated on 24/Jul/15

For l=lim_(n→+∞) a_(6n+r) ,r∈Z_6 this limit is undefined since we cannot explicitly determine which value n takes at +∞ ∀r∈Z_6 . We thus cannot be sure where along the sequence we are for any given value of r as n becomes arbitrarily large.

$${For}\:{l}=\underset{{n}\rightarrow+\infty} {\mathrm{lim}}{a}_{\mathrm{6}{n}+{r}} \:,{r}\in\mathbb{Z}_{\mathrm{6}} \:{this}\:{limit} \\ $$$${is}\:{undefined}\:{since}\:{we}\:{cannot} \\ $$$${explicitly}\:{determine}\:{which}\:{value} \\ $$$${n}\:{takes}\:{at}\:+\infty\:\forall{r}\in\mathbb{Z}_{\mathrm{6}} .\:{We}\:{thus}\: \\ $$$${cannot}\:{be}\:{sure}\:{where}\:{along}\:{the}\: \\ $$$${sequence}\:{we}\:{are}\:{for}\:{any}\:{given}\: \\ $$$${value}\:{of}\:{r}\:{as}\:{n}\:{becomes}\:{arbitrarily} \\ $$$${large}. \\ $$

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