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 yesterday


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
    yesterday










  • 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
    yesterday

















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 yesterday


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
    yesterday










  • 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
    yesterday















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













share|cite|improve this question




share|cite|improve this question








edited yesterday









Tianlalu

2,594632




2,594632










asked yesterday









RandomThinker

192




192




marked as duplicate by Nosrati, rtybase, Trevor Gunn, Arthur, user90369 yesterday


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 yesterday


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
    yesterday










  • 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
    yesterday




















  • 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
    yesterday










  • 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
    yesterday


















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
yesterday




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
yesterday












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
yesterday






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
yesterday












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 yesterday

























        answered yesterday









        Arthur

        108k7103186




        108k7103186















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