Study convergence of a recurent sequence











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How to study convergence of $(X_n)$, with $X_{n+1}={sqrt{2X_n+3}}$ and $x_1=sqrt3$ ?






sequences-and-series convergence recurrence-relations







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  • 2




    Any attempts? What have you learned so far?
    – xbh
    Nov 18 at 9:37










  • I tried to write some terms and to guess a general formula for $X_n$, but does not work
    – Raul1998
    Nov 18 at 9:42












  • try to include whatever you have in the original question itself if possible.
    – Siong Thye Goh
    Nov 18 at 9:50










  • If it is an exercice, your lecturer for sure has mentionned (one of) the fixed point theorem(s).
    – Jean Marie
    Nov 18 at 10:12















up vote
0
down vote

favorite












How to study convergence of $(X_n)$, with $X_{n+1}={sqrt{2X_n+3}}$ and $x_1=sqrt3$ ?






sequences-and-series convergence recurrence-relations







share|cite|improve this question




















  • 2




    Any attempts? What have you learned so far?
    – xbh
    Nov 18 at 9:37










  • I tried to write some terms and to guess a general formula for $X_n$, but does not work
    – Raul1998
    Nov 18 at 9:42












  • try to include whatever you have in the original question itself if possible.
    – Siong Thye Goh
    Nov 18 at 9:50










  • If it is an exercice, your lecturer for sure has mentionned (one of) the fixed point theorem(s).
    – Jean Marie
    Nov 18 at 10:12













up vote
0
down vote

favorite









up vote
0
down vote

favorite











How to study convergence of $(X_n)$, with $X_{n+1}={sqrt{2X_n+3}}$ and $x_1=sqrt3$ ?






sequences-and-series convergence recurrence-relations







share|cite|improve this question















How to study convergence of $(X_n)$, with $X_{n+1}={sqrt{2X_n+3}}$ and $x_1=sqrt3$ ?






sequences-and-series convergence recurrence-relations




sequences-and-series convergence recurrence-relations






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share|cite|improve this question













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edited Nov 18 at 10:09









Jean Marie

28.2k41848




28.2k41848










asked Nov 18 at 9:36









Raul1998

63




63








  • 2




    Any attempts? What have you learned so far?
    – xbh
    Nov 18 at 9:37










  • I tried to write some terms and to guess a general formula for $X_n$, but does not work
    – Raul1998
    Nov 18 at 9:42












  • try to include whatever you have in the original question itself if possible.
    – Siong Thye Goh
    Nov 18 at 9:50










  • If it is an exercice, your lecturer for sure has mentionned (one of) the fixed point theorem(s).
    – Jean Marie
    Nov 18 at 10:12














  • 2




    Any attempts? What have you learned so far?
    – xbh
    Nov 18 at 9:37










  • I tried to write some terms and to guess a general formula for $X_n$, but does not work
    – Raul1998
    Nov 18 at 9:42












  • try to include whatever you have in the original question itself if possible.
    – Siong Thye Goh
    Nov 18 at 9:50










  • If it is an exercice, your lecturer for sure has mentionned (one of) the fixed point theorem(s).
    – Jean Marie
    Nov 18 at 10:12








2




2




Any attempts? What have you learned so far?
– xbh
Nov 18 at 9:37




Any attempts? What have you learned so far?
– xbh
Nov 18 at 9:37












I tried to write some terms and to guess a general formula for $X_n$, but does not work
– Raul1998
Nov 18 at 9:42






I tried to write some terms and to guess a general formula for $X_n$, but does not work
– Raul1998
Nov 18 at 9:42














try to include whatever you have in the original question itself if possible.
– Siong Thye Goh
Nov 18 at 9:50




try to include whatever you have in the original question itself if possible.
– Siong Thye Goh
Nov 18 at 9:50












If it is an exercice, your lecturer for sure has mentionned (one of) the fixed point theorem(s).
– Jean Marie
Nov 18 at 10:12




If it is an exercice, your lecturer for sure has mentionned (one of) the fixed point theorem(s).
– Jean Marie
Nov 18 at 10:12










3 Answers
3






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1
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Hint:begin{align}x_{n+1}-3 &= sqrt{2x_n+3}-3 \
&=frac{2x_n+3-9}{sqrt{2x_n+3}+3} \
&=frac{2}{sqrt{2x_n+3}+3}(x_n-3)end{align}






share|cite|improve this answer





















  • I think that from here we have to find if $X_n$ is decreasing or increasing and the find the limit by crossing to the limit if recurence formula.
    – Raul1998
    Nov 18 at 10:09










  • u might want to study $|x_n-3|$.
    – Siong Thye Goh
    Nov 18 at 12:56


















up vote
0
down vote













Clearly we have $x_ngeq sqrt{3}$. Let see what is with monotonocy of the sequence $$x_{n+1}-x_{n} = sqrt{2x_n+3}-x_n = {2x_n+3-x_n^2 over sqrt{2x_n+3}+x_n }$$
We used here $$sqrt{a}-sqrt{b} ={a-b over sqrt{a}+sqrt{b} }$$
Since numerator is $>0$ we see that all depends of denumerator: $$-x_n^2+2x_n+3= -(x_n-3)(x_n+1)$$



Now since $x_n+1>0$ all depend on relation of $x_n$ with $3$ for all $n$. Now if $x_n<3$ then $$2x_n+3leq 9implies x_{n+1}< 3$$ Since $x_1< 3$ we see that $x_n< 3$ for all $n$ and so $x_{n+1}geq x_n$ for all $n$.



So our sequence is increasing and bounded in $[sqrt{3},3)$ so it is convergent.






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    Define $e_n=a_n-3$ therefore $$e_{n+1}=a_{n+1}-3=sqrt{2(e_n+3)+3}-3$$therefore $$a_{n+1}=sqrt{2e_n+9}-3={2e_nover 3+sqrt{2e_n+9}}$$and we have $$|e_{n+1}|le {2over 3}|e_n|$$ which means that $e_nto 0$ and $a_nto 3$






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      3 Answers
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      3 Answers
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      up vote
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      Hint:begin{align}x_{n+1}-3 &= sqrt{2x_n+3}-3 \
      &=frac{2x_n+3-9}{sqrt{2x_n+3}+3} \
      &=frac{2}{sqrt{2x_n+3}+3}(x_n-3)end{align}






      share|cite|improve this answer





















      • I think that from here we have to find if $X_n$ is decreasing or increasing and the find the limit by crossing to the limit if recurence formula.
        – Raul1998
        Nov 18 at 10:09










      • u might want to study $|x_n-3|$.
        – Siong Thye Goh
        Nov 18 at 12:56















      up vote
      1
      down vote













      Hint:begin{align}x_{n+1}-3 &= sqrt{2x_n+3}-3 \
      &=frac{2x_n+3-9}{sqrt{2x_n+3}+3} \
      &=frac{2}{sqrt{2x_n+3}+3}(x_n-3)end{align}






      share|cite|improve this answer





















      • I think that from here we have to find if $X_n$ is decreasing or increasing and the find the limit by crossing to the limit if recurence formula.
        – Raul1998
        Nov 18 at 10:09










      • u might want to study $|x_n-3|$.
        – Siong Thye Goh
        Nov 18 at 12:56













      up vote
      1
      down vote










      up vote
      1
      down vote









      Hint:begin{align}x_{n+1}-3 &= sqrt{2x_n+3}-3 \
      &=frac{2x_n+3-9}{sqrt{2x_n+3}+3} \
      &=frac{2}{sqrt{2x_n+3}+3}(x_n-3)end{align}






      share|cite|improve this answer












      Hint:begin{align}x_{n+1}-3 &= sqrt{2x_n+3}-3 \
      &=frac{2x_n+3-9}{sqrt{2x_n+3}+3} \
      &=frac{2}{sqrt{2x_n+3}+3}(x_n-3)end{align}







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered Nov 18 at 9:48









      Siong Thye Goh

      96.4k1462116




      96.4k1462116












      • I think that from here we have to find if $X_n$ is decreasing or increasing and the find the limit by crossing to the limit if recurence formula.
        – Raul1998
        Nov 18 at 10:09










      • u might want to study $|x_n-3|$.
        – Siong Thye Goh
        Nov 18 at 12:56


















      • I think that from here we have to find if $X_n$ is decreasing or increasing and the find the limit by crossing to the limit if recurence formula.
        – Raul1998
        Nov 18 at 10:09










      • u might want to study $|x_n-3|$.
        – Siong Thye Goh
        Nov 18 at 12:56
















      I think that from here we have to find if $X_n$ is decreasing or increasing and the find the limit by crossing to the limit if recurence formula.
      – Raul1998
      Nov 18 at 10:09




      I think that from here we have to find if $X_n$ is decreasing or increasing and the find the limit by crossing to the limit if recurence formula.
      – Raul1998
      Nov 18 at 10:09












      u might want to study $|x_n-3|$.
      – Siong Thye Goh
      Nov 18 at 12:56




      u might want to study $|x_n-3|$.
      – Siong Thye Goh
      Nov 18 at 12:56










      up vote
      0
      down vote













      Clearly we have $x_ngeq sqrt{3}$. Let see what is with monotonocy of the sequence $$x_{n+1}-x_{n} = sqrt{2x_n+3}-x_n = {2x_n+3-x_n^2 over sqrt{2x_n+3}+x_n }$$
      We used here $$sqrt{a}-sqrt{b} ={a-b over sqrt{a}+sqrt{b} }$$
      Since numerator is $>0$ we see that all depends of denumerator: $$-x_n^2+2x_n+3= -(x_n-3)(x_n+1)$$



      Now since $x_n+1>0$ all depend on relation of $x_n$ with $3$ for all $n$. Now if $x_n<3$ then $$2x_n+3leq 9implies x_{n+1}< 3$$ Since $x_1< 3$ we see that $x_n< 3$ for all $n$ and so $x_{n+1}geq x_n$ for all $n$.



      So our sequence is increasing and bounded in $[sqrt{3},3)$ so it is convergent.






      share|cite|improve this answer

























        up vote
        0
        down vote













        Clearly we have $x_ngeq sqrt{3}$. Let see what is with monotonocy of the sequence $$x_{n+1}-x_{n} = sqrt{2x_n+3}-x_n = {2x_n+3-x_n^2 over sqrt{2x_n+3}+x_n }$$
        We used here $$sqrt{a}-sqrt{b} ={a-b over sqrt{a}+sqrt{b} }$$
        Since numerator is $>0$ we see that all depends of denumerator: $$-x_n^2+2x_n+3= -(x_n-3)(x_n+1)$$



        Now since $x_n+1>0$ all depend on relation of $x_n$ with $3$ for all $n$. Now if $x_n<3$ then $$2x_n+3leq 9implies x_{n+1}< 3$$ Since $x_1< 3$ we see that $x_n< 3$ for all $n$ and so $x_{n+1}geq x_n$ for all $n$.



        So our sequence is increasing and bounded in $[sqrt{3},3)$ so it is convergent.






        share|cite|improve this answer























          up vote
          0
          down vote










          up vote
          0
          down vote









          Clearly we have $x_ngeq sqrt{3}$. Let see what is with monotonocy of the sequence $$x_{n+1}-x_{n} = sqrt{2x_n+3}-x_n = {2x_n+3-x_n^2 over sqrt{2x_n+3}+x_n }$$
          We used here $$sqrt{a}-sqrt{b} ={a-b over sqrt{a}+sqrt{b} }$$
          Since numerator is $>0$ we see that all depends of denumerator: $$-x_n^2+2x_n+3= -(x_n-3)(x_n+1)$$



          Now since $x_n+1>0$ all depend on relation of $x_n$ with $3$ for all $n$. Now if $x_n<3$ then $$2x_n+3leq 9implies x_{n+1}< 3$$ Since $x_1< 3$ we see that $x_n< 3$ for all $n$ and so $x_{n+1}geq x_n$ for all $n$.



          So our sequence is increasing and bounded in $[sqrt{3},3)$ so it is convergent.






          share|cite|improve this answer












          Clearly we have $x_ngeq sqrt{3}$. Let see what is with monotonocy of the sequence $$x_{n+1}-x_{n} = sqrt{2x_n+3}-x_n = {2x_n+3-x_n^2 over sqrt{2x_n+3}+x_n }$$
          We used here $$sqrt{a}-sqrt{b} ={a-b over sqrt{a}+sqrt{b} }$$
          Since numerator is $>0$ we see that all depends of denumerator: $$-x_n^2+2x_n+3= -(x_n-3)(x_n+1)$$



          Now since $x_n+1>0$ all depend on relation of $x_n$ with $3$ for all $n$. Now if $x_n<3$ then $$2x_n+3leq 9implies x_{n+1}< 3$$ Since $x_1< 3$ we see that $x_n< 3$ for all $n$ and so $x_{n+1}geq x_n$ for all $n$.



          So our sequence is increasing and bounded in $[sqrt{3},3)$ so it is convergent.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 18 at 14:17









          greedoid

          36.1k114591




          36.1k114591






















              up vote
              0
              down vote













              Define $e_n=a_n-3$ therefore $$e_{n+1}=a_{n+1}-3=sqrt{2(e_n+3)+3}-3$$therefore $$a_{n+1}=sqrt{2e_n+9}-3={2e_nover 3+sqrt{2e_n+9}}$$and we have $$|e_{n+1}|le {2over 3}|e_n|$$ which means that $e_nto 0$ and $a_nto 3$






              share|cite|improve this answer

























                up vote
                0
                down vote













                Define $e_n=a_n-3$ therefore $$e_{n+1}=a_{n+1}-3=sqrt{2(e_n+3)+3}-3$$therefore $$a_{n+1}=sqrt{2e_n+9}-3={2e_nover 3+sqrt{2e_n+9}}$$and we have $$|e_{n+1}|le {2over 3}|e_n|$$ which means that $e_nto 0$ and $a_nto 3$






                share|cite|improve this answer























                  up vote
                  0
                  down vote










                  up vote
                  0
                  down vote









                  Define $e_n=a_n-3$ therefore $$e_{n+1}=a_{n+1}-3=sqrt{2(e_n+3)+3}-3$$therefore $$a_{n+1}=sqrt{2e_n+9}-3={2e_nover 3+sqrt{2e_n+9}}$$and we have $$|e_{n+1}|le {2over 3}|e_n|$$ which means that $e_nto 0$ and $a_nto 3$






                  share|cite|improve this answer












                  Define $e_n=a_n-3$ therefore $$e_{n+1}=a_{n+1}-3=sqrt{2(e_n+3)+3}-3$$therefore $$a_{n+1}=sqrt{2e_n+9}-3={2e_nover 3+sqrt{2e_n+9}}$$and we have $$|e_{n+1}|le {2over 3}|e_n|$$ which means that $e_nto 0$ and $a_nto 3$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 28 at 21:02









                  Mostafa Ayaz

                  13.3k3836




                  13.3k3836






























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