Using induction to prove a sequence $a_1=1$ and $a_{n+1}=3-frac{1}{a_n}$ is increasing












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I have a sequence defined by $a_1=1$ and $a_{n+1}=3-frac{1}{a_n}$. I need to use induction to show that the sequence is increasing and $a_n<3$ for all $n$.



Also to deduce that $a_n$ is convergence and find its limit.



So far I have found: $a_n<a_{n+1}<3$ and $a_{n+1}>0$



Then $a_{n+1}=3-frac{1}{a_n}$ with $a_{n+2}=3-frac{a_n}{3a_n-1}$



But not really sure how to go any further even if this is the right way to go in the first place? Any input would be greatly be appreciated!










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    1












    $begingroup$


    I have a sequence defined by $a_1=1$ and $a_{n+1}=3-frac{1}{a_n}$. I need to use induction to show that the sequence is increasing and $a_n<3$ for all $n$.



    Also to deduce that $a_n$ is convergence and find its limit.



    So far I have found: $a_n<a_{n+1}<3$ and $a_{n+1}>0$



    Then $a_{n+1}=3-frac{1}{a_n}$ with $a_{n+2}=3-frac{a_n}{3a_n-1}$



    But not really sure how to go any further even if this is the right way to go in the first place? Any input would be greatly be appreciated!










    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      I have a sequence defined by $a_1=1$ and $a_{n+1}=3-frac{1}{a_n}$. I need to use induction to show that the sequence is increasing and $a_n<3$ for all $n$.



      Also to deduce that $a_n$ is convergence and find its limit.



      So far I have found: $a_n<a_{n+1}<3$ and $a_{n+1}>0$



      Then $a_{n+1}=3-frac{1}{a_n}$ with $a_{n+2}=3-frac{a_n}{3a_n-1}$



      But not really sure how to go any further even if this is the right way to go in the first place? Any input would be greatly be appreciated!










      share|cite|improve this question











      $endgroup$




      I have a sequence defined by $a_1=1$ and $a_{n+1}=3-frac{1}{a_n}$. I need to use induction to show that the sequence is increasing and $a_n<3$ for all $n$.



      Also to deduce that $a_n$ is convergence and find its limit.



      So far I have found: $a_n<a_{n+1}<3$ and $a_{n+1}>0$



      Then $a_{n+1}=3-frac{1}{a_n}$ with $a_{n+2}=3-frac{a_n}{3a_n-1}$



      But not really sure how to go any further even if this is the right way to go in the first place? Any input would be greatly be appreciated!







      sequences-and-series limits inequality convergence induction






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      edited Apr 20 at 12:35









      YuiTo Cheng

      2,82141139




      2,82141139










      asked Apr 20 at 11:01









      James odareJames odare

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          2 Answers
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          $begingroup$

          You have already done most of the work. By showing that $1 leq a_n < a_{n+1} < 3$ where $n in mathbb{Z}_+$, you have demonstrated that the sequence is increasing and bounded above by $3$. Now the Monotone Convergence Theorem says that any increasing sequence that is bounded above converges; indeed it converges to the supremum of ${a_n}_{n in mathbb{Z}_+}$. So you now know that $limlimits_{n to infty} a_n = alpha in mathbb{R}$ for some $$boxed{1 leq alpha leq 3}$$



          To find $alpha$, note that $f(x) = 3 - frac{1}{x}$ is a continuous real valued function for real $x > 0$. So, by continuity, and the fact that $alpha > 0$, we can put limits inside the function like so: $$limlimits_{n to infty}f(a_n) = f(limlimits_{n to infty}a_n)$$ But the LHS is just $limlimits_{n to infty}f(a_n) = limlimits_{n to infty}(3 - frac{1}{a_n}) = limlimits_{n to infty} a_{n + 1} = alpha$ while the RHS is just $f(limlimits_{n to infty}a_n) = f(alpha) = 3 - frac{1}{alpha}$. So, putting LHS $=$ RHS we get
          $$alpha = 3 - frac{1}{alpha}$$
          $$boxed{alpha^2 - 3 alpha + 1 = 0}$$ Now, solve this quadratic equation to get the value of $alpha$. You will get two values but only one of them is the answer. See if you can eliminate the wrong one using the properties of $alpha$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thank you very much!
            $endgroup$
            – James odare
            Apr 20 at 11:54










          • $begingroup$
            No problem, you're welcome.
            $endgroup$
            – ZeroXLR
            Apr 20 at 11:56



















          4












          $begingroup$

          Let $P(n)$ be the assertion “$a_ninleft[1,frac{3+sqrt5}2right)$ and $a_{n+1}>a_n$”.



          If $n=1$, then this is true, since $a_1=1$ and $a_2=2>1=a_1$.



          Let $ninmathbb N$ and assume that $P(n)$ holds. Then:





          • $a_{n+1}=3-frac1{a_n}<3$. since $a_ngeqslant1>0$;


          • $a_{n+1}geqslant1$, since $a_{n}geqslant1$ and so $3-frac1{a_n}geqslant3-frac11=2$.


          • $a_{n+2}-a_{n+1}=3-dfrac1{a_{n+1}}-a_{n+1}=dfrac{3a_{n+1}-1-{a_{n+1}}^2}{{a_{n+1}}^2}=dfrac{varphi(a_{n+1})}{{a_{n+1}}^2}$, where $varphi(x)=-x^2+3x-1$. But the roots of the quadratic polynomial $varphi(x)$ are $dfrac{3pmsqrt5}2$ and so, since $a_{n+1}$ is between them and the coefficient of $x^2$ in $varphi(x)$ is negative, $varphi(a_{n+1})>0$.


          So, $P(n+1)$ is proved.



          What I wrote above is an inductive proof of the fact that we always have $P(n)$. And it is very easy now to conclude that $lim_{ntoinfty}a_n=dfrac{3+sqrt5}2$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Great explanation thank you
            $endgroup$
            – James odare
            Apr 20 at 11:54












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          2 Answers
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          2 Answers
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          1












          $begingroup$

          You have already done most of the work. By showing that $1 leq a_n < a_{n+1} < 3$ where $n in mathbb{Z}_+$, you have demonstrated that the sequence is increasing and bounded above by $3$. Now the Monotone Convergence Theorem says that any increasing sequence that is bounded above converges; indeed it converges to the supremum of ${a_n}_{n in mathbb{Z}_+}$. So you now know that $limlimits_{n to infty} a_n = alpha in mathbb{R}$ for some $$boxed{1 leq alpha leq 3}$$



          To find $alpha$, note that $f(x) = 3 - frac{1}{x}$ is a continuous real valued function for real $x > 0$. So, by continuity, and the fact that $alpha > 0$, we can put limits inside the function like so: $$limlimits_{n to infty}f(a_n) = f(limlimits_{n to infty}a_n)$$ But the LHS is just $limlimits_{n to infty}f(a_n) = limlimits_{n to infty}(3 - frac{1}{a_n}) = limlimits_{n to infty} a_{n + 1} = alpha$ while the RHS is just $f(limlimits_{n to infty}a_n) = f(alpha) = 3 - frac{1}{alpha}$. So, putting LHS $=$ RHS we get
          $$alpha = 3 - frac{1}{alpha}$$
          $$boxed{alpha^2 - 3 alpha + 1 = 0}$$ Now, solve this quadratic equation to get the value of $alpha$. You will get two values but only one of them is the answer. See if you can eliminate the wrong one using the properties of $alpha$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thank you very much!
            $endgroup$
            – James odare
            Apr 20 at 11:54










          • $begingroup$
            No problem, you're welcome.
            $endgroup$
            – ZeroXLR
            Apr 20 at 11:56
















          1












          $begingroup$

          You have already done most of the work. By showing that $1 leq a_n < a_{n+1} < 3$ where $n in mathbb{Z}_+$, you have demonstrated that the sequence is increasing and bounded above by $3$. Now the Monotone Convergence Theorem says that any increasing sequence that is bounded above converges; indeed it converges to the supremum of ${a_n}_{n in mathbb{Z}_+}$. So you now know that $limlimits_{n to infty} a_n = alpha in mathbb{R}$ for some $$boxed{1 leq alpha leq 3}$$



          To find $alpha$, note that $f(x) = 3 - frac{1}{x}$ is a continuous real valued function for real $x > 0$. So, by continuity, and the fact that $alpha > 0$, we can put limits inside the function like so: $$limlimits_{n to infty}f(a_n) = f(limlimits_{n to infty}a_n)$$ But the LHS is just $limlimits_{n to infty}f(a_n) = limlimits_{n to infty}(3 - frac{1}{a_n}) = limlimits_{n to infty} a_{n + 1} = alpha$ while the RHS is just $f(limlimits_{n to infty}a_n) = f(alpha) = 3 - frac{1}{alpha}$. So, putting LHS $=$ RHS we get
          $$alpha = 3 - frac{1}{alpha}$$
          $$boxed{alpha^2 - 3 alpha + 1 = 0}$$ Now, solve this quadratic equation to get the value of $alpha$. You will get two values but only one of them is the answer. See if you can eliminate the wrong one using the properties of $alpha$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thank you very much!
            $endgroup$
            – James odare
            Apr 20 at 11:54










          • $begingroup$
            No problem, you're welcome.
            $endgroup$
            – ZeroXLR
            Apr 20 at 11:56














          1












          1








          1





          $begingroup$

          You have already done most of the work. By showing that $1 leq a_n < a_{n+1} < 3$ where $n in mathbb{Z}_+$, you have demonstrated that the sequence is increasing and bounded above by $3$. Now the Monotone Convergence Theorem says that any increasing sequence that is bounded above converges; indeed it converges to the supremum of ${a_n}_{n in mathbb{Z}_+}$. So you now know that $limlimits_{n to infty} a_n = alpha in mathbb{R}$ for some $$boxed{1 leq alpha leq 3}$$



          To find $alpha$, note that $f(x) = 3 - frac{1}{x}$ is a continuous real valued function for real $x > 0$. So, by continuity, and the fact that $alpha > 0$, we can put limits inside the function like so: $$limlimits_{n to infty}f(a_n) = f(limlimits_{n to infty}a_n)$$ But the LHS is just $limlimits_{n to infty}f(a_n) = limlimits_{n to infty}(3 - frac{1}{a_n}) = limlimits_{n to infty} a_{n + 1} = alpha$ while the RHS is just $f(limlimits_{n to infty}a_n) = f(alpha) = 3 - frac{1}{alpha}$. So, putting LHS $=$ RHS we get
          $$alpha = 3 - frac{1}{alpha}$$
          $$boxed{alpha^2 - 3 alpha + 1 = 0}$$ Now, solve this quadratic equation to get the value of $alpha$. You will get two values but only one of them is the answer. See if you can eliminate the wrong one using the properties of $alpha$.






          share|cite|improve this answer











          $endgroup$



          You have already done most of the work. By showing that $1 leq a_n < a_{n+1} < 3$ where $n in mathbb{Z}_+$, you have demonstrated that the sequence is increasing and bounded above by $3$. Now the Monotone Convergence Theorem says that any increasing sequence that is bounded above converges; indeed it converges to the supremum of ${a_n}_{n in mathbb{Z}_+}$. So you now know that $limlimits_{n to infty} a_n = alpha in mathbb{R}$ for some $$boxed{1 leq alpha leq 3}$$



          To find $alpha$, note that $f(x) = 3 - frac{1}{x}$ is a continuous real valued function for real $x > 0$. So, by continuity, and the fact that $alpha > 0$, we can put limits inside the function like so: $$limlimits_{n to infty}f(a_n) = f(limlimits_{n to infty}a_n)$$ But the LHS is just $limlimits_{n to infty}f(a_n) = limlimits_{n to infty}(3 - frac{1}{a_n}) = limlimits_{n to infty} a_{n + 1} = alpha$ while the RHS is just $f(limlimits_{n to infty}a_n) = f(alpha) = 3 - frac{1}{alpha}$. So, putting LHS $=$ RHS we get
          $$alpha = 3 - frac{1}{alpha}$$
          $$boxed{alpha^2 - 3 alpha + 1 = 0}$$ Now, solve this quadratic equation to get the value of $alpha$. You will get two values but only one of them is the answer. See if you can eliminate the wrong one using the properties of $alpha$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Apr 20 at 12:28

























          answered Apr 20 at 11:29









          ZeroXLRZeroXLR

          1,821620




          1,821620












          • $begingroup$
            Thank you very much!
            $endgroup$
            – James odare
            Apr 20 at 11:54










          • $begingroup$
            No problem, you're welcome.
            $endgroup$
            – ZeroXLR
            Apr 20 at 11:56


















          • $begingroup$
            Thank you very much!
            $endgroup$
            – James odare
            Apr 20 at 11:54










          • $begingroup$
            No problem, you're welcome.
            $endgroup$
            – ZeroXLR
            Apr 20 at 11:56
















          $begingroup$
          Thank you very much!
          $endgroup$
          – James odare
          Apr 20 at 11:54




          $begingroup$
          Thank you very much!
          $endgroup$
          – James odare
          Apr 20 at 11:54












          $begingroup$
          No problem, you're welcome.
          $endgroup$
          – ZeroXLR
          Apr 20 at 11:56




          $begingroup$
          No problem, you're welcome.
          $endgroup$
          – ZeroXLR
          Apr 20 at 11:56











          4












          $begingroup$

          Let $P(n)$ be the assertion “$a_ninleft[1,frac{3+sqrt5}2right)$ and $a_{n+1}>a_n$”.



          If $n=1$, then this is true, since $a_1=1$ and $a_2=2>1=a_1$.



          Let $ninmathbb N$ and assume that $P(n)$ holds. Then:





          • $a_{n+1}=3-frac1{a_n}<3$. since $a_ngeqslant1>0$;


          • $a_{n+1}geqslant1$, since $a_{n}geqslant1$ and so $3-frac1{a_n}geqslant3-frac11=2$.


          • $a_{n+2}-a_{n+1}=3-dfrac1{a_{n+1}}-a_{n+1}=dfrac{3a_{n+1}-1-{a_{n+1}}^2}{{a_{n+1}}^2}=dfrac{varphi(a_{n+1})}{{a_{n+1}}^2}$, where $varphi(x)=-x^2+3x-1$. But the roots of the quadratic polynomial $varphi(x)$ are $dfrac{3pmsqrt5}2$ and so, since $a_{n+1}$ is between them and the coefficient of $x^2$ in $varphi(x)$ is negative, $varphi(a_{n+1})>0$.


          So, $P(n+1)$ is proved.



          What I wrote above is an inductive proof of the fact that we always have $P(n)$. And it is very easy now to conclude that $lim_{ntoinfty}a_n=dfrac{3+sqrt5}2$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Great explanation thank you
            $endgroup$
            – James odare
            Apr 20 at 11:54
















          4












          $begingroup$

          Let $P(n)$ be the assertion “$a_ninleft[1,frac{3+sqrt5}2right)$ and $a_{n+1}>a_n$”.



          If $n=1$, then this is true, since $a_1=1$ and $a_2=2>1=a_1$.



          Let $ninmathbb N$ and assume that $P(n)$ holds. Then:





          • $a_{n+1}=3-frac1{a_n}<3$. since $a_ngeqslant1>0$;


          • $a_{n+1}geqslant1$, since $a_{n}geqslant1$ and so $3-frac1{a_n}geqslant3-frac11=2$.


          • $a_{n+2}-a_{n+1}=3-dfrac1{a_{n+1}}-a_{n+1}=dfrac{3a_{n+1}-1-{a_{n+1}}^2}{{a_{n+1}}^2}=dfrac{varphi(a_{n+1})}{{a_{n+1}}^2}$, where $varphi(x)=-x^2+3x-1$. But the roots of the quadratic polynomial $varphi(x)$ are $dfrac{3pmsqrt5}2$ and so, since $a_{n+1}$ is between them and the coefficient of $x^2$ in $varphi(x)$ is negative, $varphi(a_{n+1})>0$.


          So, $P(n+1)$ is proved.



          What I wrote above is an inductive proof of the fact that we always have $P(n)$. And it is very easy now to conclude that $lim_{ntoinfty}a_n=dfrac{3+sqrt5}2$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Great explanation thank you
            $endgroup$
            – James odare
            Apr 20 at 11:54














          4












          4








          4





          $begingroup$

          Let $P(n)$ be the assertion “$a_ninleft[1,frac{3+sqrt5}2right)$ and $a_{n+1}>a_n$”.



          If $n=1$, then this is true, since $a_1=1$ and $a_2=2>1=a_1$.



          Let $ninmathbb N$ and assume that $P(n)$ holds. Then:





          • $a_{n+1}=3-frac1{a_n}<3$. since $a_ngeqslant1>0$;


          • $a_{n+1}geqslant1$, since $a_{n}geqslant1$ and so $3-frac1{a_n}geqslant3-frac11=2$.


          • $a_{n+2}-a_{n+1}=3-dfrac1{a_{n+1}}-a_{n+1}=dfrac{3a_{n+1}-1-{a_{n+1}}^2}{{a_{n+1}}^2}=dfrac{varphi(a_{n+1})}{{a_{n+1}}^2}$, where $varphi(x)=-x^2+3x-1$. But the roots of the quadratic polynomial $varphi(x)$ are $dfrac{3pmsqrt5}2$ and so, since $a_{n+1}$ is between them and the coefficient of $x^2$ in $varphi(x)$ is negative, $varphi(a_{n+1})>0$.


          So, $P(n+1)$ is proved.



          What I wrote above is an inductive proof of the fact that we always have $P(n)$. And it is very easy now to conclude that $lim_{ntoinfty}a_n=dfrac{3+sqrt5}2$.






          share|cite|improve this answer









          $endgroup$



          Let $P(n)$ be the assertion “$a_ninleft[1,frac{3+sqrt5}2right)$ and $a_{n+1}>a_n$”.



          If $n=1$, then this is true, since $a_1=1$ and $a_2=2>1=a_1$.



          Let $ninmathbb N$ and assume that $P(n)$ holds. Then:





          • $a_{n+1}=3-frac1{a_n}<3$. since $a_ngeqslant1>0$;


          • $a_{n+1}geqslant1$, since $a_{n}geqslant1$ and so $3-frac1{a_n}geqslant3-frac11=2$.


          • $a_{n+2}-a_{n+1}=3-dfrac1{a_{n+1}}-a_{n+1}=dfrac{3a_{n+1}-1-{a_{n+1}}^2}{{a_{n+1}}^2}=dfrac{varphi(a_{n+1})}{{a_{n+1}}^2}$, where $varphi(x)=-x^2+3x-1$. But the roots of the quadratic polynomial $varphi(x)$ are $dfrac{3pmsqrt5}2$ and so, since $a_{n+1}$ is between them and the coefficient of $x^2$ in $varphi(x)$ is negative, $varphi(a_{n+1})>0$.


          So, $P(n+1)$ is proved.



          What I wrote above is an inductive proof of the fact that we always have $P(n)$. And it is very easy now to conclude that $lim_{ntoinfty}a_n=dfrac{3+sqrt5}2$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Apr 20 at 11:33









          José Carlos SantosJosé Carlos Santos

          177k24138251




          177k24138251












          • $begingroup$
            Great explanation thank you
            $endgroup$
            – James odare
            Apr 20 at 11:54


















          • $begingroup$
            Great explanation thank you
            $endgroup$
            – James odare
            Apr 20 at 11:54
















          $begingroup$
          Great explanation thank you
          $endgroup$
          – James odare
          Apr 20 at 11:54




          $begingroup$
          Great explanation thank you
          $endgroup$
          – James odare
          Apr 20 at 11:54


















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