How to find the $lim_{ntoinfty}s_n$ when $s_1=5, s_n =sqrt{2+s_{n-1}}$?












1












$begingroup$


How to find the $lim_{ntoinfty}s_n$ when $s_1=5, s_n =sqrt{2+s_{n-1}}$ using the Monotone Convergence Theorem?



I have the proof from my professor, but I am stuck at one step.



Proof:



Step 1: Show that it is monotonic:



Proof by induction:
Claim: $s_n > s_{n+1}, forall ninmathbb{N}$



Base case: $n=1$. $s_1 = 5 > s_2=sqrt{7}$, so it holds for $n=1$.



Assume it holds for $n=k$, now show it's true for $n=k+1$.



Assume $s_k>s_{k+1}$, then $s_{(k+1)+1} = s_{k+2} = sqrt{2+s_{k+1}} < sqrt{2+s_k} = s_{k+1}$ by assumption. Therefore, by the Principle of Mathematical Induction, the statement is true $forall ninmathbb{N}$ and ${S_n}$ is decreasing (monotonic).



Step 2: Find the limit:



It is clear $0leq s_nleq s_1 = 5forall n$, so ${S_n}$ is bounded.



Therefore, since ${S_n}$ is monotone and bounded, by the Monotone Convergence Theorem, ${S_n}$ converges.



Let $lim_{ntoinfty} s_n = L$:
begin{align}
lim_{ntoinfty}sqrt{2+s_{n-1}} &= L\
sqrt{2 + lim_{ntoinfty} s_{n-1}} &= L\
sqrt{2+L} &= L *text{This is where I get stuck...}\
2+L &= L^2\
0 &= L^2 - L - 2\
&= (L-2)(L+1)\
&implies L=-1, 2
end{align}
But $Lne-1$ since $s_ngeq 0forall n$. Therefore, $lim_{ntoinfty} s_n = 2$.



I get stuck because I am unsure why we can say $lim_{ntoinfty} s_{n-1} = L$. We said that $lim_{ntoinfty} s_{n} = L$, but nothing about the limit of $s_{n-1}$.






real-analysis limits monotone-functions







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    It is the tail that matters. The tail of $s_{n-1}$ and $s_n$ are the same. So they converge to the same thing.
    $endgroup$
    – user9077
    Dec 15 '18 at 15:55






  • 1




    $begingroup$
    What happens to $n-1$ as $ntoinfty?$
    $endgroup$
    – saulspatz
    Dec 15 '18 at 15:55
















1












$begingroup$


How to find the $lim_{ntoinfty}s_n$ when $s_1=5, s_n =sqrt{2+s_{n-1}}$ using the Monotone Convergence Theorem?



I have the proof from my professor, but I am stuck at one step.



Proof:



Step 1: Show that it is monotonic:



Proof by induction:
Claim: $s_n > s_{n+1}, forall ninmathbb{N}$



Base case: $n=1$. $s_1 = 5 > s_2=sqrt{7}$, so it holds for $n=1$.



Assume it holds for $n=k$, now show it's true for $n=k+1$.



Assume $s_k>s_{k+1}$, then $s_{(k+1)+1} = s_{k+2} = sqrt{2+s_{k+1}} < sqrt{2+s_k} = s_{k+1}$ by assumption. Therefore, by the Principle of Mathematical Induction, the statement is true $forall ninmathbb{N}$ and ${S_n}$ is decreasing (monotonic).



Step 2: Find the limit:



It is clear $0leq s_nleq s_1 = 5forall n$, so ${S_n}$ is bounded.



Therefore, since ${S_n}$ is monotone and bounded, by the Monotone Convergence Theorem, ${S_n}$ converges.



Let $lim_{ntoinfty} s_n = L$:
begin{align}
lim_{ntoinfty}sqrt{2+s_{n-1}} &= L\
sqrt{2 + lim_{ntoinfty} s_{n-1}} &= L\
sqrt{2+L} &= L *text{This is where I get stuck...}\
2+L &= L^2\
0 &= L^2 - L - 2\
&= (L-2)(L+1)\
&implies L=-1, 2
end{align}
But $Lne-1$ since $s_ngeq 0forall n$. Therefore, $lim_{ntoinfty} s_n = 2$.



I get stuck because I am unsure why we can say $lim_{ntoinfty} s_{n-1} = L$. We said that $lim_{ntoinfty} s_{n} = L$, but nothing about the limit of $s_{n-1}$.






real-analysis limits monotone-functions







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    It is the tail that matters. The tail of $s_{n-1}$ and $s_n$ are the same. So they converge to the same thing.
    $endgroup$
    – user9077
    Dec 15 '18 at 15:55






  • 1




    $begingroup$
    What happens to $n-1$ as $ntoinfty?$
    $endgroup$
    – saulspatz
    Dec 15 '18 at 15:55














1












1








1





$begingroup$


How to find the $lim_{ntoinfty}s_n$ when $s_1=5, s_n =sqrt{2+s_{n-1}}$ using the Monotone Convergence Theorem?



I have the proof from my professor, but I am stuck at one step.



Proof:



Step 1: Show that it is monotonic:



Proof by induction:
Claim: $s_n > s_{n+1}, forall ninmathbb{N}$



Base case: $n=1$. $s_1 = 5 > s_2=sqrt{7}$, so it holds for $n=1$.



Assume it holds for $n=k$, now show it's true for $n=k+1$.



Assume $s_k>s_{k+1}$, then $s_{(k+1)+1} = s_{k+2} = sqrt{2+s_{k+1}} < sqrt{2+s_k} = s_{k+1}$ by assumption. Therefore, by the Principle of Mathematical Induction, the statement is true $forall ninmathbb{N}$ and ${S_n}$ is decreasing (monotonic).



Step 2: Find the limit:



It is clear $0leq s_nleq s_1 = 5forall n$, so ${S_n}$ is bounded.



Therefore, since ${S_n}$ is monotone and bounded, by the Monotone Convergence Theorem, ${S_n}$ converges.



Let $lim_{ntoinfty} s_n = L$:
begin{align}
lim_{ntoinfty}sqrt{2+s_{n-1}} &= L\
sqrt{2 + lim_{ntoinfty} s_{n-1}} &= L\
sqrt{2+L} &= L *text{This is where I get stuck...}\
2+L &= L^2\
0 &= L^2 - L - 2\
&= (L-2)(L+1)\
&implies L=-1, 2
end{align}
But $Lne-1$ since $s_ngeq 0forall n$. Therefore, $lim_{ntoinfty} s_n = 2$.



I get stuck because I am unsure why we can say $lim_{ntoinfty} s_{n-1} = L$. We said that $lim_{ntoinfty} s_{n} = L$, but nothing about the limit of $s_{n-1}$.






real-analysis limits monotone-functions







share|cite|improve this question











$endgroup$




How to find the $lim_{ntoinfty}s_n$ when $s_1=5, s_n =sqrt{2+s_{n-1}}$ using the Monotone Convergence Theorem?



I have the proof from my professor, but I am stuck at one step.



Proof:



Step 1: Show that it is monotonic:



Proof by induction:
Claim: $s_n > s_{n+1}, forall ninmathbb{N}$



Base case: $n=1$. $s_1 = 5 > s_2=sqrt{7}$, so it holds for $n=1$.



Assume it holds for $n=k$, now show it's true for $n=k+1$.



Assume $s_k>s_{k+1}$, then $s_{(k+1)+1} = s_{k+2} = sqrt{2+s_{k+1}} < sqrt{2+s_k} = s_{k+1}$ by assumption. Therefore, by the Principle of Mathematical Induction, the statement is true $forall ninmathbb{N}$ and ${S_n}$ is decreasing (monotonic).



Step 2: Find the limit:



It is clear $0leq s_nleq s_1 = 5forall n$, so ${S_n}$ is bounded.



Therefore, since ${S_n}$ is monotone and bounded, by the Monotone Convergence Theorem, ${S_n}$ converges.



Let $lim_{ntoinfty} s_n = L$:
begin{align}
lim_{ntoinfty}sqrt{2+s_{n-1}} &= L\
sqrt{2 + lim_{ntoinfty} s_{n-1}} &= L\
sqrt{2+L} &= L *text{This is where I get stuck...}\
2+L &= L^2\
0 &= L^2 - L - 2\
&= (L-2)(L+1)\
&implies L=-1, 2
end{align}
But $Lne-1$ since $s_ngeq 0forall n$. Therefore, $lim_{ntoinfty} s_n = 2$.



I get stuck because I am unsure why we can say $lim_{ntoinfty} s_{n-1} = L$. We said that $lim_{ntoinfty} s_{n} = L$, but nothing about the limit of $s_{n-1}$.






real-analysis limits monotone-functions




real-analysis limits monotone-functions






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 15 '18 at 15:58









Bernard

119k639112




119k639112










asked Dec 15 '18 at 15:50









kaisakaisa

1019




1019








  • 1




    $begingroup$
    It is the tail that matters. The tail of $s_{n-1}$ and $s_n$ are the same. So they converge to the same thing.
    $endgroup$
    – user9077
    Dec 15 '18 at 15:55






  • 1




    $begingroup$
    What happens to $n-1$ as $ntoinfty?$
    $endgroup$
    – saulspatz
    Dec 15 '18 at 15:55














  • 1




    $begingroup$
    It is the tail that matters. The tail of $s_{n-1}$ and $s_n$ are the same. So they converge to the same thing.
    $endgroup$
    – user9077
    Dec 15 '18 at 15:55






  • 1




    $begingroup$
    What happens to $n-1$ as $ntoinfty?$
    $endgroup$
    – saulspatz
    Dec 15 '18 at 15:55








1




1




$begingroup$
It is the tail that matters. The tail of $s_{n-1}$ and $s_n$ are the same. So they converge to the same thing.
$endgroup$
– user9077
Dec 15 '18 at 15:55




$begingroup$
It is the tail that matters. The tail of $s_{n-1}$ and $s_n$ are the same. So they converge to the same thing.
$endgroup$
– user9077
Dec 15 '18 at 15:55




1




1




$begingroup$
What happens to $n-1$ as $ntoinfty?$
$endgroup$
– saulspatz
Dec 15 '18 at 15:55




$begingroup$
What happens to $n-1$ as $ntoinfty?$
$endgroup$
– saulspatz
Dec 15 '18 at 15:55










2 Answers
2






active

oldest

votes


















2












$begingroup$

Write down the definition of a limit to show that if $(a_n)$ is any converging sequence, then $(a_{n-1})$ is also a converging sequence and both limits are the same.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    The sequence is decreasing, because $s_{n+1}<s_n$ is equivalent to
    $$
    s_n+2<s_n^2
    $$
    which is satisfied when $s_n<-1$ or $s_n>2$. The latter is true by induction: $s_1=5>2$; suppose $s_k>2$; then $s_{k+1}=sqrt{s_k+2}>sqrt{2+2}=2$.



    Your proof is good as well.



    The sequence is bounded below by $0$, hence it converges. If $l$ is its limit, then $lge0$ and
    $$
    l=lim_{ntoinfty}s_n=lim_{ntoinfty}s_{n+1}=sqrt{l+2}
    $$
    Therefore $l=2$.



    Actually, $lim_{ntoinfty}s_{n+1}$ should really be $lim_{ntoinfty}s'_n$, where
    $$
    s'_n=s_{n+1}
    $$
    It's essentially going through the definition of limit to show that $(s_n)$ converges if and only if $(s'_n)$ and, in this case, they have the same limit.



    Since $s'_n=sqrt{s_n+1}$, the theorems on limits also allow us to say that $$l=lim_{ntoinfty}s'_n=lim_{ntoinfty}sqrt{s_n+2}=sqrt{l+2}$$



    Note: I wouldn't use $s_{n-1}$, but it's just personal preference.






    share|cite|improve this answer









    $endgroup$













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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2












      $begingroup$

      Write down the definition of a limit to show that if $(a_n)$ is any converging sequence, then $(a_{n-1})$ is also a converging sequence and both limits are the same.






      share|cite|improve this answer









      $endgroup$


















        2












        $begingroup$

        Write down the definition of a limit to show that if $(a_n)$ is any converging sequence, then $(a_{n-1})$ is also a converging sequence and both limits are the same.






        share|cite|improve this answer









        $endgroup$
















          2












          2








          2





          $begingroup$

          Write down the definition of a limit to show that if $(a_n)$ is any converging sequence, then $(a_{n-1})$ is also a converging sequence and both limits are the same.






          share|cite|improve this answer









          $endgroup$



          Write down the definition of a limit to show that if $(a_n)$ is any converging sequence, then $(a_{n-1})$ is also a converging sequence and both limits are the same.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 15 '18 at 15:52









          MindlackMindlack

          2,81717




          2,81717























              0












              $begingroup$

              The sequence is decreasing, because $s_{n+1}<s_n$ is equivalent to
              $$
              s_n+2<s_n^2
              $$
              which is satisfied when $s_n<-1$ or $s_n>2$. The latter is true by induction: $s_1=5>2$; suppose $s_k>2$; then $s_{k+1}=sqrt{s_k+2}>sqrt{2+2}=2$.



              Your proof is good as well.



              The sequence is bounded below by $0$, hence it converges. If $l$ is its limit, then $lge0$ and
              $$
              l=lim_{ntoinfty}s_n=lim_{ntoinfty}s_{n+1}=sqrt{l+2}
              $$
              Therefore $l=2$.



              Actually, $lim_{ntoinfty}s_{n+1}$ should really be $lim_{ntoinfty}s'_n$, where
              $$
              s'_n=s_{n+1}
              $$
              It's essentially going through the definition of limit to show that $(s_n)$ converges if and only if $(s'_n)$ and, in this case, they have the same limit.



              Since $s'_n=sqrt{s_n+1}$, the theorems on limits also allow us to say that $$l=lim_{ntoinfty}s'_n=lim_{ntoinfty}sqrt{s_n+2}=sqrt{l+2}$$



              Note: I wouldn't use $s_{n-1}$, but it's just personal preference.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                The sequence is decreasing, because $s_{n+1}<s_n$ is equivalent to
                $$
                s_n+2<s_n^2
                $$
                which is satisfied when $s_n<-1$ or $s_n>2$. The latter is true by induction: $s_1=5>2$; suppose $s_k>2$; then $s_{k+1}=sqrt{s_k+2}>sqrt{2+2}=2$.



                Your proof is good as well.



                The sequence is bounded below by $0$, hence it converges. If $l$ is its limit, then $lge0$ and
                $$
                l=lim_{ntoinfty}s_n=lim_{ntoinfty}s_{n+1}=sqrt{l+2}
                $$
                Therefore $l=2$.



                Actually, $lim_{ntoinfty}s_{n+1}$ should really be $lim_{ntoinfty}s'_n$, where
                $$
                s'_n=s_{n+1}
                $$
                It's essentially going through the definition of limit to show that $(s_n)$ converges if and only if $(s'_n)$ and, in this case, they have the same limit.



                Since $s'_n=sqrt{s_n+1}$, the theorems on limits also allow us to say that $$l=lim_{ntoinfty}s'_n=lim_{ntoinfty}sqrt{s_n+2}=sqrt{l+2}$$



                Note: I wouldn't use $s_{n-1}$, but it's just personal preference.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  The sequence is decreasing, because $s_{n+1}<s_n$ is equivalent to
                  $$
                  s_n+2<s_n^2
                  $$
                  which is satisfied when $s_n<-1$ or $s_n>2$. The latter is true by induction: $s_1=5>2$; suppose $s_k>2$; then $s_{k+1}=sqrt{s_k+2}>sqrt{2+2}=2$.



                  Your proof is good as well.



                  The sequence is bounded below by $0$, hence it converges. If $l$ is its limit, then $lge0$ and
                  $$
                  l=lim_{ntoinfty}s_n=lim_{ntoinfty}s_{n+1}=sqrt{l+2}
                  $$
                  Therefore $l=2$.



                  Actually, $lim_{ntoinfty}s_{n+1}$ should really be $lim_{ntoinfty}s'_n$, where
                  $$
                  s'_n=s_{n+1}
                  $$
                  It's essentially going through the definition of limit to show that $(s_n)$ converges if and only if $(s'_n)$ and, in this case, they have the same limit.



                  Since $s'_n=sqrt{s_n+1}$, the theorems on limits also allow us to say that $$l=lim_{ntoinfty}s'_n=lim_{ntoinfty}sqrt{s_n+2}=sqrt{l+2}$$



                  Note: I wouldn't use $s_{n-1}$, but it's just personal preference.






                  share|cite|improve this answer









                  $endgroup$



                  The sequence is decreasing, because $s_{n+1}<s_n$ is equivalent to
                  $$
                  s_n+2<s_n^2
                  $$
                  which is satisfied when $s_n<-1$ or $s_n>2$. The latter is true by induction: $s_1=5>2$; suppose $s_k>2$; then $s_{k+1}=sqrt{s_k+2}>sqrt{2+2}=2$.



                  Your proof is good as well.



                  The sequence is bounded below by $0$, hence it converges. If $l$ is its limit, then $lge0$ and
                  $$
                  l=lim_{ntoinfty}s_n=lim_{ntoinfty}s_{n+1}=sqrt{l+2}
                  $$
                  Therefore $l=2$.



                  Actually, $lim_{ntoinfty}s_{n+1}$ should really be $lim_{ntoinfty}s'_n$, where
                  $$
                  s'_n=s_{n+1}
                  $$
                  It's essentially going through the definition of limit to show that $(s_n)$ converges if and only if $(s'_n)$ and, in this case, they have the same limit.



                  Since $s'_n=sqrt{s_n+1}$, the theorems on limits also allow us to say that $$l=lim_{ntoinfty}s'_n=lim_{ntoinfty}sqrt{s_n+2}=sqrt{l+2}$$



                  Note: I wouldn't use $s_{n-1}$, but it's just personal preference.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 15 '18 at 16:08









                  egregegreg

                  180k1485202




                  180k1485202






























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