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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-




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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