Fix $a > 0$ and let $x_1 > sqrt a$ [duplicate]











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  • Find the limit if it exists of $S_{n+1} = frac{1}{2}(S_n +frac{A}{S_n})$

    2 answers




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My current thought is to take $x_{n+1}-x_n$ and to prove the difference is smaller than $sqrt a$. But it won't help me to prove $x_n$ converges to $sqrt a$.






real-analysis convergence recurrence-relations







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marked as duplicate by Nosrati, rtybase, Trevor Gunn, Arthur, user90369 Nov 19 at 20:14


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  • Well, the question starts by asking you to show that the sequence is decreasing; so $x_{n + 1} - x_n < sqrt{a}$ is useless, because you actually need to show that $x_{n + 1} - x_n le 0$. Then you'll probably invoke a theorem about monotone convergence.
    – T. Bongers
    Nov 19 at 16:30










  • I answered here to this question sometimes ago, in a different context (it was required that the approximating succession ${x_n}_{ninmathbb{N}}$ was made of only rational numbers, but the method of proof is basically the same).
    – Daniele Tampieri
    Nov 19 at 16:31

















up vote
0
down vote

favorite













This question already has an answer here:




  • Find the limit if it exists of $S_{n+1} = frac{1}{2}(S_n +frac{A}{S_n})$

    2 answers




enter image description here



My current thought is to take $x_{n+1}-x_n$ and to prove the difference is smaller than $sqrt a$. But it won't help me to prove $x_n$ converges to $sqrt a$.






real-analysis convergence recurrence-relations







share|cite|improve this question















marked as duplicate by Nosrati, rtybase, Trevor Gunn, Arthur, user90369 Nov 19 at 20:14


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.















  • Well, the question starts by asking you to show that the sequence is decreasing; so $x_{n + 1} - x_n < sqrt{a}$ is useless, because you actually need to show that $x_{n + 1} - x_n le 0$. Then you'll probably invoke a theorem about monotone convergence.
    – T. Bongers
    Nov 19 at 16:30










  • I answered here to this question sometimes ago, in a different context (it was required that the approximating succession ${x_n}_{ninmathbb{N}}$ was made of only rational numbers, but the method of proof is basically the same).
    – Daniele Tampieri
    Nov 19 at 16:31















up vote
0
down vote

favorite









up vote
0
down vote

favorite












This question already has an answer here:




  • Find the limit if it exists of $S_{n+1} = frac{1}{2}(S_n +frac{A}{S_n})$

    2 answers




enter image description here



My current thought is to take $x_{n+1}-x_n$ and to prove the difference is smaller than $sqrt a$. But it won't help me to prove $x_n$ converges to $sqrt a$.






real-analysis convergence recurrence-relations







share|cite|improve this question
















This question already has an answer here:




  • Find the limit if it exists of $S_{n+1} = frac{1}{2}(S_n +frac{A}{S_n})$

    2 answers




enter image description here



My current thought is to take $x_{n+1}-x_n$ and to prove the difference is smaller than $sqrt a$. But it won't help me to prove $x_n$ converges to $sqrt a$.





This question already has an answer here:




  • Find the limit if it exists of $S_{n+1} = frac{1}{2}(S_n +frac{A}{S_n})$

    2 answers







real-analysis convergence recurrence-relations




real-analysis convergence recurrence-relations






share|cite|improve this question















share|cite|improve this question













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edited Nov 19 at 16:32









Tianlalu

2,9361935




2,9361935










asked Nov 19 at 16:21









RandomThinker

192




192




marked as duplicate by Nosrati, rtybase, Trevor Gunn, Arthur, user90369 Nov 19 at 20:14


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.






marked as duplicate by Nosrati, rtybase, Trevor Gunn, Arthur, user90369 Nov 19 at 20:14


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.














  • Well, the question starts by asking you to show that the sequence is decreasing; so $x_{n + 1} - x_n < sqrt{a}$ is useless, because you actually need to show that $x_{n + 1} - x_n le 0$. Then you'll probably invoke a theorem about monotone convergence.
    – T. Bongers
    Nov 19 at 16:30










  • I answered here to this question sometimes ago, in a different context (it was required that the approximating succession ${x_n}_{ninmathbb{N}}$ was made of only rational numbers, but the method of proof is basically the same).
    – Daniele Tampieri
    Nov 19 at 16:31




















  • Well, the question starts by asking you to show that the sequence is decreasing; so $x_{n + 1} - x_n < sqrt{a}$ is useless, because you actually need to show that $x_{n + 1} - x_n le 0$. Then you'll probably invoke a theorem about monotone convergence.
    – T. Bongers
    Nov 19 at 16:30










  • I answered here to this question sometimes ago, in a different context (it was required that the approximating succession ${x_n}_{ninmathbb{N}}$ was made of only rational numbers, but the method of proof is basically the same).
    – Daniele Tampieri
    Nov 19 at 16:31


















Well, the question starts by asking you to show that the sequence is decreasing; so $x_{n + 1} - x_n < sqrt{a}$ is useless, because you actually need to show that $x_{n + 1} - x_n le 0$. Then you'll probably invoke a theorem about monotone convergence.
– T. Bongers
Nov 19 at 16:30




Well, the question starts by asking you to show that the sequence is decreasing; so $x_{n + 1} - x_n < sqrt{a}$ is useless, because you actually need to show that $x_{n + 1} - x_n le 0$. Then you'll probably invoke a theorem about monotone convergence.
– T. Bongers
Nov 19 at 16:30












I answered here to this question sometimes ago, in a different context (it was required that the approximating succession ${x_n}_{ninmathbb{N}}$ was made of only rational numbers, but the method of proof is basically the same).
– Daniele Tampieri
Nov 19 at 16:31






I answered here to this question sometimes ago, in a different context (it was required that the approximating succession ${x_n}_{ninmathbb{N}}$ was made of only rational numbers, but the method of proof is basically the same).
– Daniele Tampieri
Nov 19 at 16:31












1 Answer
1






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Assuming $x_1>sqrtalpha$, show that $x_{n+1}>sqrtalpha$. Then use that to show $x_{n+1}-x_n<0$, which proves the sequence is monotonically decreasing.



Therefore (it is monotonically decreasing and bounded below) the sequence has a limit $L$. The limit must satisfy:
$$
L=frac12left(L+fracalpha Lright)
$$
which means you can figure out exactly what it is.






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    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    2
    down vote













    Assuming $x_1>sqrtalpha$, show that $x_{n+1}>sqrtalpha$. Then use that to show $x_{n+1}-x_n<0$, which proves the sequence is monotonically decreasing.



    Therefore (it is monotonically decreasing and bounded below) the sequence has a limit $L$. The limit must satisfy:
    $$
    L=frac12left(L+fracalpha Lright)
    $$
    which means you can figure out exactly what it is.






    share|cite|improve this answer



























      up vote
      2
      down vote













      Assuming $x_1>sqrtalpha$, show that $x_{n+1}>sqrtalpha$. Then use that to show $x_{n+1}-x_n<0$, which proves the sequence is monotonically decreasing.



      Therefore (it is monotonically decreasing and bounded below) the sequence has a limit $L$. The limit must satisfy:
      $$
      L=frac12left(L+fracalpha Lright)
      $$
      which means you can figure out exactly what it is.






      share|cite|improve this answer

























        up vote
        2
        down vote










        up vote
        2
        down vote









        Assuming $x_1>sqrtalpha$, show that $x_{n+1}>sqrtalpha$. Then use that to show $x_{n+1}-x_n<0$, which proves the sequence is monotonically decreasing.



        Therefore (it is monotonically decreasing and bounded below) the sequence has a limit $L$. The limit must satisfy:
        $$
        L=frac12left(L+fracalpha Lright)
        $$
        which means you can figure out exactly what it is.






        share|cite|improve this answer














        Assuming $x_1>sqrtalpha$, show that $x_{n+1}>sqrtalpha$. Then use that to show $x_{n+1}-x_n<0$, which proves the sequence is monotonically decreasing.



        Therefore (it is monotonically decreasing and bounded below) the sequence has a limit $L$. The limit must satisfy:
        $$
        L=frac12left(L+fracalpha Lright)
        $$
        which means you can figure out exactly what it is.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Nov 19 at 16:39

























        answered Nov 19 at 16:32









        Arthur

        109k7103186




        109k7103186















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