Understanding the proof: If $a_1=1$ and $a_n+1=sqrta_1+a_2+cdots+a_n.$ Then, $limlimits_ntoinftyfraca_nn=frac12$

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I want to understand a section in the following proof, let $a_1=1$ and $$a_n+1=sqrta_1+a_2+cdots+a_n,$$ then, $$limlimits_ntoinftyfraca_nn=frac12.$$



Here is my question: the above question was solved in Let $a_n+1=sqrta_1+a_2+cdots+a_n$ .Prove that $ limlimits_n rightarrow infty fraca_nn=frac12$, by taking that $a_n+1^2=a_n^2+a_n.$ Please, can you show me how $a_n+1^2=a_n^2+a_n$ was gotten?




real-analysis sequences-and-series analysis




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  • The linked question is different. It deals with a square root; your's has the nth root.
    – user58697
    Jul 15 at 23:43










  • Well, you can begin with $a_n+1^2 = a_1 + cdots + a_n$ and $a_n^2 = a_1 + cdots + a_n-1$.
    – Sangchul Lee
    Jul 15 at 23:45










  • @ user58697 and spaceisdarkgreen: Sorry it was a typo! I've corrected it!
    – Mike
    Jul 15 at 23:45










  • @Mike they don't assume it. They prove it
    – mathworker21
    Jul 15 at 23:50










  • @mathworker21: Sorry for the language misuse!
    – Mike
    Jul 16 at 0:52














up vote
1
down vote

favorite












I want to understand a section in the following proof, let $a_1=1$ and $$a_n+1=sqrta_1+a_2+cdots+a_n,$$ then, $$limlimits_ntoinftyfraca_nn=frac12.$$



Here is my question: the above question was solved in Let $a_n+1=sqrta_1+a_2+cdots+a_n$ .Prove that $ limlimits_n rightarrow infty fraca_nn=frac12$, by taking that $a_n+1^2=a_n^2+a_n.$ Please, can you show me how $a_n+1^2=a_n^2+a_n$ was gotten?




real-analysis sequences-and-series analysis




share|cite|improve this question





















  • The linked question is different. It deals with a square root; your's has the nth root.
    – user58697
    Jul 15 at 23:43










  • Well, you can begin with $a_n+1^2 = a_1 + cdots + a_n$ and $a_n^2 = a_1 + cdots + a_n-1$.
    – Sangchul Lee
    Jul 15 at 23:45










  • @ user58697 and spaceisdarkgreen: Sorry it was a typo! I've corrected it!
    – Mike
    Jul 15 at 23:45










  • @Mike they don't assume it. They prove it
    – mathworker21
    Jul 15 at 23:50










  • @mathworker21: Sorry for the language misuse!
    – Mike
    Jul 16 at 0:52












up vote
1
down vote

favorite









up vote
1
down vote

favorite











I want to understand a section in the following proof, let $a_1=1$ and $$a_n+1=sqrta_1+a_2+cdots+a_n,$$ then, $$limlimits_ntoinftyfraca_nn=frac12.$$



Here is my question: the above question was solved in Let $a_n+1=sqrta_1+a_2+cdots+a_n$ .Prove that $ limlimits_n rightarrow infty fraca_nn=frac12$, by taking that $a_n+1^2=a_n^2+a_n.$ Please, can you show me how $a_n+1^2=a_n^2+a_n$ was gotten?




real-analysis sequences-and-series analysis




share|cite|improve this question













I want to understand a section in the following proof, let $a_1=1$ and $$a_n+1=sqrta_1+a_2+cdots+a_n,$$ then, $$limlimits_ntoinftyfraca_nn=frac12.$$



Here is my question: the above question was solved in Let $a_n+1=sqrta_1+a_2+cdots+a_n$ .Prove that $ limlimits_n rightarrow infty fraca_nn=frac12$, by taking that $a_n+1^2=a_n^2+a_n.$ Please, can you show me how $a_n+1^2=a_n^2+a_n$ was gotten?





real-analysis sequences-and-series analysis





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Jul 16 at 2:00
























asked Jul 15 at 23:39









Mike

74512




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  • The linked question is different. It deals with a square root; your's has the nth root.
    – user58697
    Jul 15 at 23:43










  • Well, you can begin with $a_n+1^2 = a_1 + cdots + a_n$ and $a_n^2 = a_1 + cdots + a_n-1$.
    – Sangchul Lee
    Jul 15 at 23:45










  • @ user58697 and spaceisdarkgreen: Sorry it was a typo! I've corrected it!
    – Mike
    Jul 15 at 23:45










  • @Mike they don't assume it. They prove it
    – mathworker21
    Jul 15 at 23:50










  • @mathworker21: Sorry for the language misuse!
    – Mike
    Jul 16 at 0:52
















  • The linked question is different. It deals with a square root; your's has the nth root.
    – user58697
    Jul 15 at 23:43










  • Well, you can begin with $a_n+1^2 = a_1 + cdots + a_n$ and $a_n^2 = a_1 + cdots + a_n-1$.
    – Sangchul Lee
    Jul 15 at 23:45










  • @ user58697 and spaceisdarkgreen: Sorry it was a typo! I've corrected it!
    – Mike
    Jul 15 at 23:45










  • @Mike they don't assume it. They prove it
    – mathworker21
    Jul 15 at 23:50










  • @mathworker21: Sorry for the language misuse!
    – Mike
    Jul 16 at 0:52















The linked question is different. It deals with a square root; your's has the nth root.
– user58697
Jul 15 at 23:43




The linked question is different. It deals with a square root; your's has the nth root.
– user58697
Jul 15 at 23:43












Well, you can begin with $a_n+1^2 = a_1 + cdots + a_n$ and $a_n^2 = a_1 + cdots + a_n-1$.
– Sangchul Lee
Jul 15 at 23:45




Well, you can begin with $a_n+1^2 = a_1 + cdots + a_n$ and $a_n^2 = a_1 + cdots + a_n-1$.
– Sangchul Lee
Jul 15 at 23:45












@ user58697 and spaceisdarkgreen: Sorry it was a typo! I've corrected it!
– Mike
Jul 15 at 23:45




@ user58697 and spaceisdarkgreen: Sorry it was a typo! I've corrected it!
– Mike
Jul 15 at 23:45












@Mike they don't assume it. They prove it
– mathworker21
Jul 15 at 23:50




@Mike they don't assume it. They prove it
– mathworker21
Jul 15 at 23:50












@mathworker21: Sorry for the language misuse!
– Mike
Jul 16 at 0:52




@mathworker21: Sorry for the language misuse!
– Mike
Jul 16 at 0:52










1 Answer
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5
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accepted










Note that $$a_n+1^2 = a_1+ldots +a_n-1 + a_n= left(sqrta_1+ldots +a_n-1right)^2 + a_n= a_n^2+a_n$$






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  • I shouldn't get a downvote! I only asked to get clarifications, please!
    – Mike
    Jul 16 at 2:01










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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
5
down vote



accepted










Note that $$a_n+1^2 = a_1+ldots +a_n-1 + a_n= left(sqrta_1+ldots +a_n-1right)^2 + a_n= a_n^2+a_n$$






share|cite|improve this answer























  • I shouldn't get a downvote! I only asked to get clarifications, please!
    – Mike
    Jul 16 at 2:01














up vote
5
down vote



accepted










Note that $$a_n+1^2 = a_1+ldots +a_n-1 + a_n= left(sqrta_1+ldots +a_n-1right)^2 + a_n= a_n^2+a_n$$






share|cite|improve this answer























  • I shouldn't get a downvote! I only asked to get clarifications, please!
    – Mike
    Jul 16 at 2:01












up vote
5
down vote



accepted







up vote
5
down vote



accepted






Note that $$a_n+1^2 = a_1+ldots +a_n-1 + a_n= left(sqrta_1+ldots +a_n-1right)^2 + a_n= a_n^2+a_n$$






share|cite|improve this answer















Note that $$a_n+1^2 = a_1+ldots +a_n-1 + a_n= left(sqrta_1+ldots +a_n-1right)^2 + a_n= a_n^2+a_n$$







share|cite|improve this answer















share|cite|improve this answer



share|cite|improve this answer








edited Jul 16 at 16:55









amWhy

189k25219431




189k25219431











answered Jul 15 at 23:46









spaceisdarkgreen

27.6k21547




27.6k21547











  • I shouldn't get a downvote! I only asked to get clarifications, please!
    – Mike
    Jul 16 at 2:01
















  • I shouldn't get a downvote! I only asked to get clarifications, please!
    – Mike
    Jul 16 at 2:01















I shouldn't get a downvote! I only asked to get clarifications, please!
– Mike
Jul 16 at 2:01




I shouldn't get a downvote! I only asked to get clarifications, please!
– Mike
Jul 16 at 2:01












 

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