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Question Number 4597 by FilupSmith last updated on 10/Feb/16

a_n =(√a_(n−1) ) a_1 =c Solve: S = lim_(n→∞) a_n

$${a}_{{n}} =\sqrt{{a}_{{n}−\mathrm{1}} } \\ $$$${a}_{\mathrm{1}} ={c} \\ $$$$ \\ $$$$\mathrm{Solve}: \\ $$$${S}\:=\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:{a}_{{n}} \\ $$

Commented by FilupSmith last updated on 10/Feb/16

Would a solution be: for c>0 a_1 =c^(1/2) a_2 =(c^(1/2) )^(1/2) =(c^(1/2^2 ) ) a_3 =(c^(1/2^2 ) )^(1/2) =c^(1/2^3 ) a_n =c^(1/2^n ) ∴ as n→∞ a_∞ =c^(1/∞) =c^0 a_∞ =1 ∴ lim_(n→∞) a_n = 1 for c=0 lim_(n→∞) a_n = 0

$$\mathrm{Would}\:\mathrm{a}\:\mathrm{solution}\:\mathrm{be}: \\ $$$$ \\ $$$$\mathrm{for}\:{c}>\mathrm{0} \\ $$$${a}_{\mathrm{1}} ={c}^{\mathrm{1}/\mathrm{2}} \\ $$$${a}_{\mathrm{2}} =\left({c}^{\mathrm{1}/\mathrm{2}} \right)^{\mathrm{1}/\mathrm{2}} \\ $$$$=\left({c}^{\mathrm{1}/\mathrm{2}^{\mathrm{2}} } \right) \\ $$$${a}_{\mathrm{3}} =\left({c}^{\mathrm{1}/\mathrm{2}^{\mathrm{2}} } \right)^{\mathrm{1}/\mathrm{2}} \\ $$$$={c}^{\mathrm{1}/\mathrm{2}^{\mathrm{3}} } \\ $$$$ \\ $$$${a}_{{n}} ={c}^{\frac{\mathrm{1}}{\mathrm{2}^{{n}} }} \\ $$$$\therefore\:{as}\:{n}\rightarrow\infty \\ $$$${a}_{\infty} ={c}^{\mathrm{1}/\infty} ={c}^{\mathrm{0}} \\ $$$${a}_{\infty} =\mathrm{1} \\ $$$$ \\ $$$$\therefore\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:{a}_{{n}} =\:\mathrm{1} \\ $$$$ \\ $$$$\mathrm{for}\:{c}=\mathrm{0} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:{a}_{{n}} =\:\mathrm{0} \\ $$

Commented by Yozzii last updated on 10/Feb/16

This appears to be legitimate for if c=0, a_n =0^(1/2^(n−1) ) =0 ∀n∈N⇒lim_(n→+∞) a_n =lim_(n→+∞) 0=0. The term a_∞ does not exist by the way. It′s better to say ′ as n→∞, 2^(n−1) →∞ ⇒(1/2^(n−1) )→0⇒c^(1/2^(n−1) ) →c^0 =1⇒a_n →1.′ To further convince your audience of the obvious pattern that a_n =c^(1/2^(n−1) ) , given that a_1 =c>0 and a_(n+1) =(√a_n ), use induction on n. (I saw that you wrote a_1 =c^(1/2) when the information gave a_1 =c. It′s really a_2 =(√a_1 )=c^(1/2) ⇒ a_3 =(√a_2 )=c^(1/2^2 ) ⇒a_4 =(√a_3 )=c^(1/2^3 ) ⇒...⇒a_n =c^(1/2^(n−1) ) (n∈N).)

$${This}\:{appears}\:{to}\:{be}\:{legitimate}\:{for}\:{if} \\ $$$${c}=\mathrm{0},\:{a}_{{n}} =\mathrm{0}^{\mathrm{1}/\mathrm{2}^{{n}−\mathrm{1}} } =\mathrm{0}\:\:\forall{n}\in\mathbb{N}\Rightarrow\underset{{n}\rightarrow+\infty} {\mathrm{lim}}{a}_{{n}} =\underset{{n}\rightarrow+\infty} {\mathrm{lim}0}=\mathrm{0}. \\ $$$${The}\:{term}\:{a}_{\infty} \:{does}\:{not}\:{exist}\:{by}\:{the}\:{way}. \\ $$$${It}'{s}\:{better}\:{to}\:{say}\:'\:{as}\:{n}\rightarrow\infty,\:\mathrm{2}^{{n}−\mathrm{1}} \rightarrow\infty \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{2}^{{n}−\mathrm{1}} }\rightarrow\mathrm{0}\Rightarrow{c}^{\mathrm{1}/\mathrm{2}^{{n}−\mathrm{1}} } \rightarrow{c}^{\mathrm{0}} =\mathrm{1}\Rightarrow{a}_{{n}} \rightarrow\mathrm{1}.' \\ $$$${To}\:{further}\:{convince}\:{your}\:{audience} \\ $$$${of}\:{the}\:{obvious}\:{pattern}\:{that}\:{a}_{{n}} ={c}^{\mathrm{1}/\mathrm{2}^{{n}−\mathrm{1}} } , \\ $$$${given}\:{that}\:{a}_{\mathrm{1}} ={c}>\mathrm{0}\:{and}\:{a}_{{n}+\mathrm{1}} =\sqrt{{a}_{{n}} }, \\ $$$${use}\:{induction}\:{on}\:{n}.\:\left({I}\:{saw}\:{that}\:{you}\right. \\ $$$${wrote}\:{a}_{\mathrm{1}} ={c}^{\mathrm{1}/\mathrm{2}} \:{when}\:{the}\:{information} \\ $$$${gave}\:{a}_{\mathrm{1}} ={c}.\:{It}'{s}\:{really}\:{a}_{\mathrm{2}} =\sqrt{{a}_{\mathrm{1}} }={c}^{\mathrm{1}/\mathrm{2}} \Rightarrow \\ $$$$\left.{a}_{\mathrm{3}} =\sqrt{{a}_{\mathrm{2}} }={c}^{\mathrm{1}/\mathrm{2}^{\mathrm{2}} } \Rightarrow{a}_{\mathrm{4}} =\sqrt{{a}_{\mathrm{3}} }={c}^{\mathrm{1}/\mathrm{2}^{\mathrm{3}} } \Rightarrow...\Rightarrow{a}_{{n}} ={c}^{\mathrm{1}/\mathrm{2}^{{n}−\mathrm{1}} } \:\:\left({n}\in\mathbb{N}\right).\right) \\ $$$$ \\ $$

Commented by FilupSmith last updated on 11/Feb/16

Thanks. I do not know induction. I have been meaning to learn but i can′t find an easy to understand explanation

$$\mathrm{Thanks}.\:\mathrm{I}\:\mathrm{do}\:\mathrm{not}\:\mathrm{know}\:\mathrm{induction}.\:\mathrm{I}\:\mathrm{have} \\ $$$$\mathrm{been}\:\mathrm{meaning}\:\mathrm{to}\:\mathrm{learn}\:\mathrm{but}\:\mathrm{i}\:\mathrm{can}'\mathrm{t} \\ $$$$\mathrm{find}\:\mathrm{an}\:\mathrm{easy}\:\mathrm{to}\:\mathrm{understand}\:\mathrm{explanation} \\ $$

Commented by Yozzii last updated on 11/Feb/16

The underlying principle of mathematical induction is the idea that you can obtain an integer k+1 if you know what k is,being an integer. So, given an integer k you can get the next integer by adding 1 to k⇒k+1. So, since you know what your initial integer is, e.g 1, you can find any integer greater than (or less than) 1 if you continue on from 1 to 2 ,then 2 to 3 ,then 3 to 4, ...., then k to k+1,then....to infinity. You can assign a mathematical idea P that depends on n to use induction to show that P(n) is true for all integers n since from, say P(1), you can deduce P(k),P(k+1),P(k+2)....

$${The}\:{underlying}\:{principle}\:{of}\: \\ $$$${mathematical}\:{induction}\:{is}\: \\ $$$${the}\:{idea}\:{that}\:{you}\:{can}\:{obtain}\:{an}\:{integer} \\ $$$${k}+\mathrm{1}\:{if}\:{you}\:{know}\:{what}\:{k}\:{is},{being}\:{an} \\ $$$${integer}.\:{So},\:{given}\:{an}\:{integer}\:{k}\: \\ $$$${you}\:{can}\:{get}\:{the}\:{next}\:{integer}\:{by}\:{adding} \\ $$$$\mathrm{1}\:{to}\:{k}\Rightarrow{k}+\mathrm{1}.\:{So},\:{since}\:{you}\:{know}\: \\ $$$${what}\:{your}\:{initial}\:{integer}\:{is},\:{e}.{g}\:\mathrm{1}, \\ $$$${you}\:{can}\:{find}\:{any}\:{integer}\:{greater}\:{than} \\ $$$$\left({or}\:{less}\:{than}\right)\:\mathrm{1}\:{if}\:{you}\:{continue}\:{on} \\ $$$${from}\:\mathrm{1}\:{to}\:\mathrm{2}\:,{then}\:\mathrm{2}\:{to}\:\mathrm{3}\:,{then}\:\mathrm{3}\:{to}\:\mathrm{4}, \\ $$$$....,\:{then}\:{k}\:{to}\:{k}+\mathrm{1},{then}....{to}\:{infinity}. \\ $$$${You}\:{can}\:{assign}\:{a}\:{mathematical} \\ $$$${idea}\:{P}\:\:{that}\:{depends}\:{on}\:{n}\:{to}\:{use}\: \\ $$$${induction}\:{to}\:{show}\:{that}\:{P}\left({n}\right)\:{is}\:{true} \\ $$$${for}\:{all}\:{integers}\:{n}\:{since}\:{from},\:{say}\:{P}\left(\mathrm{1}\right), \\ $$$${you}\:{can}\:{deduce}\:{P}\left({k}\right),{P}\left({k}+\mathrm{1}\right),{P}\left({k}+\mathrm{2}\right).... \\ $$$$ \\ $$$$ \\ $$

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