Showing that the following sequence converges and finding its limit.

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The sequence is given by:
$$a_1 = 3/2 , a_n+1 = sqrt 3a_n - 2$$



1-I tried to find the limit and calculations lead me to either the limit $L = 2$ or $L =1$, so then how can I choose between them?



2- For studying monotonicity of this sequence I got:



$$a_n+1 - a_n = frac3(a_n - a_n-1)a_n+1 + a_n $$



Then how can I know that the value is greater than or less than zero?



3-Finally how can I study boundedness?






calculus sequences-and-series







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    up vote
    -1
    down vote

    favorite












    The sequence is given by:
    $$a_1 = 3/2 , a_n+1 = sqrt 3a_n - 2$$



    1-I tried to find the limit and calculations lead me to either the limit $L = 2$ or $L =1$, so then how can I choose between them?



    2- For studying monotonicity of this sequence I got:



    $$a_n+1 - a_n = frac3(a_n - a_n-1)a_n+1 + a_n $$



    Then how can I know that the value is greater than or less than zero?



    3-Finally how can I study boundedness?






    calculus sequences-and-series







    share|cite|improve this question























      up vote
      -1
      down vote

      favorite









      up vote
      -1
      down vote

      favorite











      The sequence is given by:
      $$a_1 = 3/2 , a_n+1 = sqrt 3a_n - 2$$



      1-I tried to find the limit and calculations lead me to either the limit $L = 2$ or $L =1$, so then how can I choose between them?



      2- For studying monotonicity of this sequence I got:



      $$a_n+1 - a_n = frac3(a_n - a_n-1)a_n+1 + a_n $$



      Then how can I know that the value is greater than or less than zero?



      3-Finally how can I study boundedness?






      calculus sequences-and-series







      share|cite|improve this question













      The sequence is given by:
      $$a_1 = 3/2 , a_n+1 = sqrt 3a_n - 2$$



      1-I tried to find the limit and calculations lead me to either the limit $L = 2$ or $L =1$, so then how can I choose between them?



      2- For studying monotonicity of this sequence I got:



      $$a_n+1 - a_n = frac3(a_n - a_n-1)a_n+1 + a_n $$



      Then how can I know that the value is greater than or less than zero?



      3-Finally how can I study boundedness?






      calculus sequences-and-series




      calculus sequences-and-series






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      asked 2 days ago







      user591644



























          2 Answers
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          Observe that $a_1<a_2$. In general, show that $a_n-1<a_n$
          implies $sqrt3a_n-1-2<sqrt3a_n-2$ etc.



          Also $a_1<2$. Show that $a_n<2$ implies $sqrt3a_n-2<2$ etc.






          share|cite|improve this answer




















          • what about the limit?
            – user591644
            2 days ago






          • 2




            If it's an increasing sequence, and starts at 1.5? @Idonotknow
            – Lord Shark the Unknown
            2 days ago

















          up vote
          0
          down vote













          HINT



          We have that for



          $$1<a_n<2 implies 1<sqrt 3a_n - 2<2$$



          therefore the sequence is bounded, moreover we have



          $$a_n+1-a_n = sqrt 3a_n - 2-a_n=(sqrt 3a_n - 2-a_n)fracsqrt 3a_n - 2+a_nsqrt 3a_n - 2+a_n=$$$$=frac-a_n^2+3a_n-2sqrt 3a_n - 2+a_n=frac-(a_n-1)(a_n-2)sqrt 3a_n - 2+a_n>0$$






          share|cite|improve this answer




















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            2 Answers
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            2 Answers
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            up vote
            2
            down vote













            Observe that $a_1<a_2$. In general, show that $a_n-1<a_n$
            implies $sqrt3a_n-1-2<sqrt3a_n-2$ etc.



            Also $a_1<2$. Show that $a_n<2$ implies $sqrt3a_n-2<2$ etc.






            share|cite|improve this answer




















            • what about the limit?
              – user591644
              2 days ago






            • 2




              If it's an increasing sequence, and starts at 1.5? @Idonotknow
              – Lord Shark the Unknown
              2 days ago














            up vote
            2
            down vote













            Observe that $a_1<a_2$. In general, show that $a_n-1<a_n$
            implies $sqrt3a_n-1-2<sqrt3a_n-2$ etc.



            Also $a_1<2$. Show that $a_n<2$ implies $sqrt3a_n-2<2$ etc.






            share|cite|improve this answer




















            • what about the limit?
              – user591644
              2 days ago






            • 2




              If it's an increasing sequence, and starts at 1.5? @Idonotknow
              – Lord Shark the Unknown
              2 days ago












            up vote
            2
            down vote










            up vote
            2
            down vote









            Observe that $a_1<a_2$. In general, show that $a_n-1<a_n$
            implies $sqrt3a_n-1-2<sqrt3a_n-2$ etc.



            Also $a_1<2$. Show that $a_n<2$ implies $sqrt3a_n-2<2$ etc.






            share|cite|improve this answer












            Observe that $a_1<a_2$. In general, show that $a_n-1<a_n$
            implies $sqrt3a_n-1-2<sqrt3a_n-2$ etc.



            Also $a_1<2$. Show that $a_n<2$ implies $sqrt3a_n-2<2$ etc.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 2 days ago









            Lord Shark the Unknown

            88.8k955115




            88.8k955115











            • what about the limit?
              – user591644
              2 days ago






            • 2




              If it's an increasing sequence, and starts at 1.5? @Idonotknow
              – Lord Shark the Unknown
              2 days ago
















            • what about the limit?
              – user591644
              2 days ago






            • 2




              If it's an increasing sequence, and starts at 1.5? @Idonotknow
              – Lord Shark the Unknown
              2 days ago















            what about the limit?
            – user591644
            2 days ago




            what about the limit?
            – user591644
            2 days ago




            2




            2




            If it's an increasing sequence, and starts at 1.5? @Idonotknow
            – Lord Shark the Unknown
            2 days ago




            If it's an increasing sequence, and starts at 1.5? @Idonotknow
            – Lord Shark the Unknown
            2 days ago










            up vote
            0
            down vote













            HINT



            We have that for



            $$1<a_n<2 implies 1<sqrt 3a_n - 2<2$$



            therefore the sequence is bounded, moreover we have



            $$a_n+1-a_n = sqrt 3a_n - 2-a_n=(sqrt 3a_n - 2-a_n)fracsqrt 3a_n - 2+a_nsqrt 3a_n - 2+a_n=$$$$=frac-a_n^2+3a_n-2sqrt 3a_n - 2+a_n=frac-(a_n-1)(a_n-2)sqrt 3a_n - 2+a_n>0$$






            share|cite|improve this answer
























              up vote
              0
              down vote













              HINT



              We have that for



              $$1<a_n<2 implies 1<sqrt 3a_n - 2<2$$



              therefore the sequence is bounded, moreover we have



              $$a_n+1-a_n = sqrt 3a_n - 2-a_n=(sqrt 3a_n - 2-a_n)fracsqrt 3a_n - 2+a_nsqrt 3a_n - 2+a_n=$$$$=frac-a_n^2+3a_n-2sqrt 3a_n - 2+a_n=frac-(a_n-1)(a_n-2)sqrt 3a_n - 2+a_n>0$$






              share|cite|improve this answer






















                up vote
                0
                down vote










                up vote
                0
                down vote









                HINT



                We have that for



                $$1<a_n<2 implies 1<sqrt 3a_n - 2<2$$



                therefore the sequence is bounded, moreover we have



                $$a_n+1-a_n = sqrt 3a_n - 2-a_n=(sqrt 3a_n - 2-a_n)fracsqrt 3a_n - 2+a_nsqrt 3a_n - 2+a_n=$$$$=frac-a_n^2+3a_n-2sqrt 3a_n - 2+a_n=frac-(a_n-1)(a_n-2)sqrt 3a_n - 2+a_n>0$$






                share|cite|improve this answer












                HINT



                We have that for



                $$1<a_n<2 implies 1<sqrt 3a_n - 2<2$$



                therefore the sequence is bounded, moreover we have



                $$a_n+1-a_n = sqrt 3a_n - 2-a_n=(sqrt 3a_n - 2-a_n)fracsqrt 3a_n - 2+a_nsqrt 3a_n - 2+a_n=$$$$=frac-a_n^2+3a_n-2sqrt 3a_n - 2+a_n=frac-(a_n-1)(a_n-2)sqrt 3a_n - 2+a_n>0$$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 2 days ago









                gimusi

                71.4k73786




                71.4k73786



























                     

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