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Can someone help me with proving the convergence of a sequence written in this form? [duplicate]
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This question already has an answer here:
Evaluating the limit of a sequence given by recurrence relation $a_1=sqrt2$, $a_n+1=sqrt2+a_n$. Is my solution correct? [duplicate]
2 answers
From this exercise:
If $s_1 = sqrt2$, and $$s_n+1 = sqrt2+sqrts_n quad (n = 1,2,3,dots),$$ prove that $s_n$ converges, and that $s_n< 2$ for $n = 1,2,3,dots$.
in the book Principles of Mathematical Analysis, I find this book very obscure about many concepts, so I really need some help.
Can someone help me with proving the convergence of a sequence written in this form?
real-analysis sequences-and-series
edited Jul 19 at 10:49
BDN
573417
asked Jul 19 at 7:19
Miles123K
165
marked as duplicate by Claude Leibovici, rtybase, Did, Strants, Xander Henderson Jul 19 at 15:54
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.
add a comment |Â
up vote
2
down vote
favorite
This question already has an answer here:
Evaluating the limit of a sequence given by recurrence relation $a_1=sqrt2$, $a_n+1=sqrt2+a_n$. Is my solution correct? [duplicate]
2 answers
From this exercise:
If $s_1 = sqrt2$, and $$s_n+1 = sqrt2+sqrts_n quad (n = 1,2,3,dots),$$ prove that $s_n$ converges, and that $s_n< 2$ for $n = 1,2,3,dots$.
in the book Principles of Mathematical Analysis, I find this book very obscure about many concepts, so I really need some help.
Can someone help me with proving the convergence of a sequence written in this form?
real-analysis sequences-and-series
edited Jul 19 at 10:49
BDN
573417
asked Jul 19 at 7:19
Miles123K
165
marked as duplicate by Claude Leibovici, rtybase, Did, Strants, Xander Henderson Jul 19 at 15:54
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.
You can use Mathjax : math.meta.stackexchange.com/questions/5020/…
– nicomezi
Jul 19 at 7:22
How about this one math.stackexchange.com/questions/2702895/…?
– rtybase
Jul 19 at 8:00
@rtybase This one answers the second part of the question. Thanks for pointing that out!
– Miles123K
Jul 19 at 8:05
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
This question already has an answer here:
Evaluating the limit of a sequence given by recurrence relation $a_1=sqrt2$, $a_n+1=sqrt2+a_n$. Is my solution correct? [duplicate]
2 answers
From this exercise:
If $s_1 = sqrt2$, and $$s_n+1 = sqrt2+sqrts_n quad (n = 1,2,3,dots),$$ prove that $s_n$ converges, and that $s_n< 2$ for $n = 1,2,3,dots$.
in the book Principles of Mathematical Analysis, I find this book very obscure about many concepts, so I really need some help.
Can someone help me with proving the convergence of a sequence written in this form?
real-analysis sequences-and-series
edited Jul 19 at 10:49
BDN
573417
asked Jul 19 at 7:19
Miles123K
165
This question already has an answer here:
Evaluating the limit of a sequence given by recurrence relation $a_1=sqrt2$, $a_n+1=sqrt2+a_n$. Is my solution correct? [duplicate]
2 answers
From this exercise:
If $s_1 = sqrt2$, and $$s_n+1 = sqrt2+sqrts_n quad (n = 1,2,3,dots),$$ prove that $s_n$ converges, and that $s_n< 2$ for $n = 1,2,3,dots$.
in the book Principles of Mathematical Analysis, I find this book very obscure about many concepts, so I really need some help.
Can someone help me with proving the convergence of a sequence written in this form?
This question already has an answer here:
Evaluating the limit of a sequence given by recurrence relation $a_1=sqrt2$, $a_n+1=sqrt2+a_n$. Is my solution correct? [duplicate]
2 answers
real-analysis sequences-and-series
edited Jul 19 at 10:49
BDN
573417
asked Jul 19 at 7:19
Miles123K
165
edited Jul 19 at 10:49
BDN
573417
edited Jul 19 at 10:49
BDN
573417
edited Jul 19 at 10:49
BDN
573417
573417
asked Jul 19 at 7:19
Miles123K
165
asked Jul 19 at 7:19
Miles123K
165
asked Jul 19 at 7:19
Miles123K
165
165
marked as duplicate by Claude Leibovici, rtybase, Did, Strants, Xander Henderson Jul 19 at 15:54
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 Claude Leibovici, rtybase, Did, Strants, Xander Henderson Jul 19 at 15:54
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.
You can use Mathjax : math.meta.stackexchange.com/questions/5020/…
– nicomezi
Jul 19 at 7:22
How about this one math.stackexchange.com/questions/2702895/…?
– rtybase
Jul 19 at 8:00
@rtybase This one answers the second part of the question. Thanks for pointing that out!
– Miles123K
Jul 19 at 8:05
add a comment |Â
You can use Mathjax : math.meta.stackexchange.com/questions/5020/…
– nicomezi
Jul 19 at 7:22
How about this one math.stackexchange.com/questions/2702895/…?
– rtybase
Jul 19 at 8:00
@rtybase This one answers the second part of the question. Thanks for pointing that out!
– Miles123K
Jul 19 at 8:05
You can use Mathjax : math.meta.stackexchange.com/questions/5020/…
– nicomezi
Jul 19 at 7:22
You can use Mathjax : math.meta.stackexchange.com/questions/5020/…
– nicomezi
Jul 19 at 7:22
How about this one math.stackexchange.com/questions/2702895/…?
– rtybase
Jul 19 at 8:00
How about this one math.stackexchange.com/questions/2702895/…?
– rtybase
Jul 19 at 8:00
@rtybase This one answers the second part of the question. Thanks for pointing that out!
– Miles123K
Jul 19 at 8:05
@rtybase This one answers the second part of the question. Thanks for pointing that out!
– Miles123K
Jul 19 at 8:05
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
2
down vote
accepted
If such a limit exists we must have $$l=sqrt2+sqrt l$$or $$l^2=2+sqrt l$$define $e_n=s_n-l$. We show that $e_nto 0$. We have $$s_n+1=sqrt2+sqrts_n$$therefore $$e_n+1=sqrt2+sqrts_n-l\=dfrac2+sqrts_n-l^2sqrt2+sqrts_n+l\=dfrac2+sqrts_n-l^2sqrt2+sqrts_n+l\=dfracsqrts_n-sqrt lsqrt2+sqrts_n+l\=dfrace_n(sqrt2+sqrts_n+l)(sqrt s_n+sqrt l)$$therefore $$|e_n+1|=|dfrace_n(sqrt2+sqrts_n+l)(sqrt s_n+sqrt l)|=dfrac(sqrt2+sqrts_n+l)(sqrt s_n+sqrt l)le dfraclsqrt lledfrac(lsqrt l)^n$$since both $l$ and $sqrt l$ are non-negative and $$sqrt2+sqrts_n+l>l\sqrt s_n+sqrt l>sqrt l$$also the last inequality is attained by iteratively applying the one before that i.e.$$|e_n+1|<dfraclsqrt l<dfrac(lsqrt l)^2<cdots<dfrac(lsqrt l)^n$$ which means that $|e_n|to 0$ or $e_nto 0$
answered Jul 19 at 7:33
Mostafa Ayaz
8,6023630
Wait can you elaborate on the last step? How does e-sub-n approach 0?
– Miles123K
Jul 19 at 7:57
Sure! I clarified that in the context
– Mostafa Ayaz
Jul 19 at 8:04
Thank you so much! Now I understand.
– Miles123K
Jul 19 at 8:08
You're welcome :)
– Mostafa Ayaz
Jul 19 at 8:20
add a comment |Â
up vote
2
down vote
Show by indzction, that $s_n<2$ for all $n$.
Show by induction that $s_n+1 > s_n$ for all $n$.
From 1. and 2. it follows then , that $(s_n)$ is convergent.
answered Jul 19 at 7:24
Fred
37.4k1237
$1$ and $2$ imply the covergence of $(s_n)$ by monotone convergence?
– Atif Farooq
Jul 19 at 7:40
Yes, the monotone convergence theorem shows that the sequence is convergent.
– Fred
Jul 19 at 7:58
add a comment |Â
up vote
0
down vote
Because nobody mentioned this method, not even in the question I flagged as duplicate ...
Proposition 1. $0<s_n <2$
By induction. $0<s_1=sqrt2<2$. Let's assume (induction hypothesis)
$$0<s_n<2 Rightarrow
0<sqrts_n<sqrt2Rightarrow \
0<2<2+sqrts_n<2+sqrt2<4 Rightarrow \
0<sqrt2+sqrts_n<2 Rightarrow \
0<s_n+1<2$$
Proposition 2. $s_n$ is ascending.
We have $s_n+1=f(s_n)$ where $f(x)=sqrt2+sqrtx$. But $f'(x)=frac14 sqrt2 + sqrtx sqrtx>0$, so $f(x)$ is ascending for $x>0$. We also have that $s_2=sqrt2+sqrtsqrt2>sqrt2=s_1$, thus
$$f(s_2)geq f(s_1) iff s_3geq s_2$$
and by induction
$$f(s_n)geq f(s_n-1) iff s_n+1geq s_n$$
As a result, we have a monotone and bounded sequence, so it converges ... to the solution of
$$L=sqrt2+sqrtL$$
answered Jul 19 at 8:22
rtybase
8,86721433
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
If such a limit exists we must have $$l=sqrt2+sqrt l$$or $$l^2=2+sqrt l$$define $e_n=s_n-l$. We show that $e_nto 0$. We have $$s_n+1=sqrt2+sqrts_n$$therefore $$e_n+1=sqrt2+sqrts_n-l\=dfrac2+sqrts_n-l^2sqrt2+sqrts_n+l\=dfrac2+sqrts_n-l^2sqrt2+sqrts_n+l\=dfracsqrts_n-sqrt lsqrt2+sqrts_n+l\=dfrace_n(sqrt2+sqrts_n+l)(sqrt s_n+sqrt l)$$therefore $$|e_n+1|=|dfrace_n(sqrt2+sqrts_n+l)(sqrt s_n+sqrt l)|=dfrac(sqrt2+sqrts_n+l)(sqrt s_n+sqrt l)le dfraclsqrt lledfrac(lsqrt l)^n$$since both $l$ and $sqrt l$ are non-negative and $$sqrt2+sqrts_n+l>l\sqrt s_n+sqrt l>sqrt l$$also the last inequality is attained by iteratively applying the one before that i.e.$$|e_n+1|<dfraclsqrt l<dfrac(lsqrt l)^2<cdots<dfrac(lsqrt l)^n$$ which means that $|e_n|to 0$ or $e_nto 0$
answered Jul 19 at 7:33
Mostafa Ayaz
8,6023630
Wait can you elaborate on the last step? How does e-sub-n approach 0?
– Miles123K
Jul 19 at 7:57
Sure! I clarified that in the context
– Mostafa Ayaz
Jul 19 at 8:04
Thank you so much! Now I understand.
– Miles123K
Jul 19 at 8:08
You're welcome :)
– Mostafa Ayaz
Jul 19 at 8:20
add a comment |Â
up vote
2
down vote
accepted
If such a limit exists we must have $$l=sqrt2+sqrt l$$or $$l^2=2+sqrt l$$define $e_n=s_n-l$. We show that $e_nto 0$. We have $$s_n+1=sqrt2+sqrts_n$$therefore $$e_n+1=sqrt2+sqrts_n-l\=dfrac2+sqrts_n-l^2sqrt2+sqrts_n+l\=dfrac2+sqrts_n-l^2sqrt2+sqrts_n+l\=dfracsqrts_n-sqrt lsqrt2+sqrts_n+l\=dfrace_n(sqrt2+sqrts_n+l)(sqrt s_n+sqrt l)$$therefore $$|e_n+1|=|dfrace_n(sqrt2+sqrts_n+l)(sqrt s_n+sqrt l)|=dfrac(sqrt2+sqrts_n+l)(sqrt s_n+sqrt l)le dfraclsqrt lledfrac(lsqrt l)^n$$since both $l$ and $sqrt l$ are non-negative and $$sqrt2+sqrts_n+l>l\sqrt s_n+sqrt l>sqrt l$$also the last inequality is attained by iteratively applying the one before that i.e.$$|e_n+1|<dfraclsqrt l<dfrac(lsqrt l)^2<cdots<dfrac(lsqrt l)^n$$ which means that $|e_n|to 0$ or $e_nto 0$
answered Jul 19 at 7:33
Mostafa Ayaz
8,6023630
Wait can you elaborate on the last step? How does e-sub-n approach 0?
– Miles123K
Jul 19 at 7:57
Sure! I clarified that in the context
– Mostafa Ayaz
Jul 19 at 8:04
Thank you so much! Now I understand.
– Miles123K
Jul 19 at 8:08
You're welcome :)
– Mostafa Ayaz
Jul 19 at 8:20
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
If such a limit exists we must have $$l=sqrt2+sqrt l$$or $$l^2=2+sqrt l$$define $e_n=s_n-l$. We show that $e_nto 0$. We have $$s_n+1=sqrt2+sqrts_n$$therefore $$e_n+1=sqrt2+sqrts_n-l\=dfrac2+sqrts_n-l^2sqrt2+sqrts_n+l\=dfrac2+sqrts_n-l^2sqrt2+sqrts_n+l\=dfracsqrts_n-sqrt lsqrt2+sqrts_n+l\=dfrace_n(sqrt2+sqrts_n+l)(sqrt s_n+sqrt l)$$therefore $$|e_n+1|=|dfrace_n(sqrt2+sqrts_n+l)(sqrt s_n+sqrt l)|=dfrac(sqrt2+sqrts_n+l)(sqrt s_n+sqrt l)le dfraclsqrt lledfrac(lsqrt l)^n$$since both $l$ and $sqrt l$ are non-negative and $$sqrt2+sqrts_n+l>l\sqrt s_n+sqrt l>sqrt l$$also the last inequality is attained by iteratively applying the one before that i.e.$$|e_n+1|<dfraclsqrt l<dfrac(lsqrt l)^2<cdots<dfrac(lsqrt l)^n$$ which means that $|e_n|to 0$ or $e_nto 0$
answered Jul 19 at 7:33
Mostafa Ayaz
8,6023630
If such a limit exists we must have $$l=sqrt2+sqrt l$$or $$l^2=2+sqrt l$$define $e_n=s_n-l$. We show that $e_nto 0$. We have $$s_n+1=sqrt2+sqrts_n$$therefore $$e_n+1=sqrt2+sqrts_n-l\=dfrac2+sqrts_n-l^2sqrt2+sqrts_n+l\=dfrac2+sqrts_n-l^2sqrt2+sqrts_n+l\=dfracsqrts_n-sqrt lsqrt2+sqrts_n+l\=dfrace_n(sqrt2+sqrts_n+l)(sqrt s_n+sqrt l)$$therefore $$|e_n+1|=|dfrace_n(sqrt2+sqrts_n+l)(sqrt s_n+sqrt l)|=dfrac(sqrt2+sqrts_n+l)(sqrt s_n+sqrt l)le dfraclsqrt lledfrac(lsqrt l)^n$$since both $l$ and $sqrt l$ are non-negative and $$sqrt2+sqrts_n+l>l\sqrt s_n+sqrt l>sqrt l$$also the last inequality is attained by iteratively applying the one before that i.e.$$|e_n+1|<dfraclsqrt l<dfrac(lsqrt l)^2<cdots<dfrac(lsqrt l)^n$$ which means that $|e_n|to 0$ or $e_nto 0$
answered Jul 19 at 7:33
Mostafa Ayaz
8,6023630
edited Jul 19 at 8:03
answered Jul 19 at 7:33
Mostafa Ayaz
8,6023630
answered Jul 19 at 7:33
Mostafa Ayaz
8,6023630
answered Jul 19 at 7:33
Mostafa Ayaz
8,6023630
8,6023630
Wait can you elaborate on the last step? How does e-sub-n approach 0?
– Miles123K
Jul 19 at 7:57
Sure! I clarified that in the context
– Mostafa Ayaz
Jul 19 at 8:04
Thank you so much! Now I understand.
– Miles123K
Jul 19 at 8:08
You're welcome :)
– Mostafa Ayaz
Jul 19 at 8:20
add a comment |Â
Wait can you elaborate on the last step? How does e-sub-n approach 0?
– Miles123K
Jul 19 at 7:57
Sure! I clarified that in the context
– Mostafa Ayaz
Jul 19 at 8:04
Thank you so much! Now I understand.
– Miles123K
Jul 19 at 8:08
You're welcome :)
– Mostafa Ayaz
Jul 19 at 8:20
Wait can you elaborate on the last step? How does e-sub-n approach 0?
– Miles123K
Jul 19 at 7:57
Wait can you elaborate on the last step? How does e-sub-n approach 0?
– Miles123K
Jul 19 at 7:57
Sure! I clarified that in the context
– Mostafa Ayaz
Jul 19 at 8:04
Sure! I clarified that in the context
– Mostafa Ayaz
Jul 19 at 8:04
Thank you so much! Now I understand.
– Miles123K
Jul 19 at 8:08
Thank you so much! Now I understand.
– Miles123K
Jul 19 at 8:08
You're welcome :)
– Mostafa Ayaz
Jul 19 at 8:20
You're welcome :)
– Mostafa Ayaz
Jul 19 at 8:20
add a comment |Â
up vote
2
down vote
Show by indzction, that $s_n<2$ for all $n$.
Show by induction that $s_n+1 > s_n$ for all $n$.
From 1. and 2. it follows then , that $(s_n)$ is convergent.
answered Jul 19 at 7:24
Fred
37.4k1237
$1$ and $2$ imply the covergence of $(s_n)$ by monotone convergence?
– Atif Farooq
Jul 19 at 7:40
Yes, the monotone convergence theorem shows that the sequence is convergent.
– Fred
Jul 19 at 7:58
add a comment |Â
up vote
2
down vote
Show by indzction, that $s_n<2$ for all $n$.
Show by induction that $s_n+1 > s_n$ for all $n$.
From 1. and 2. it follows then , that $(s_n)$ is convergent.
answered Jul 19 at 7:24
Fred
37.4k1237
$1$ and $2$ imply the covergence of $(s_n)$ by monotone convergence?
– Atif Farooq
Jul 19 at 7:40
Yes, the monotone convergence theorem shows that the sequence is convergent.
– Fred
Jul 19 at 7:58
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Show by indzction, that $s_n<2$ for all $n$.
Show by induction that $s_n+1 > s_n$ for all $n$.
From 1. and 2. it follows then , that $(s_n)$ is convergent.
answered Jul 19 at 7:24
Fred
37.4k1237
Show by indzction, that $s_n<2$ for all $n$.
Show by induction that $s_n+1 > s_n$ for all $n$.
From 1. and 2. it follows then , that $(s_n)$ is convergent.
answered Jul 19 at 7:24
Fred
37.4k1237
answered Jul 19 at 7:24
Fred
37.4k1237
answered Jul 19 at 7:24
Fred
37.4k1237
answered Jul 19 at 7:24
Fred
37.4k1237
37.4k1237
$1$ and $2$ imply the covergence of $(s_n)$ by monotone convergence?
– Atif Farooq
Jul 19 at 7:40
Yes, the monotone convergence theorem shows that the sequence is convergent.
– Fred
Jul 19 at 7:58
add a comment |Â
$1$ and $2$ imply the covergence of $(s_n)$ by monotone convergence?
– Atif Farooq
Jul 19 at 7:40
Yes, the monotone convergence theorem shows that the sequence is convergent.
– Fred
Jul 19 at 7:58
$1$ and $2$ imply the covergence of $(s_n)$ by monotone convergence?
– Atif Farooq
Jul 19 at 7:40
$1$ and $2$ imply the covergence of $(s_n)$ by monotone convergence?
– Atif Farooq
Jul 19 at 7:40
Yes, the monotone convergence theorem shows that the sequence is convergent.
– Fred
Jul 19 at 7:58
Yes, the monotone convergence theorem shows that the sequence is convergent.
– Fred
Jul 19 at 7:58
add a comment |Â
up vote
0
down vote
Because nobody mentioned this method, not even in the question I flagged as duplicate ...
Proposition 1. $0<s_n <2$
By induction. $0<s_1=sqrt2<2$. Let's assume (induction hypothesis)
$$0<s_n<2 Rightarrow
0<sqrts_n<sqrt2Rightarrow \
0<2<2+sqrts_n<2+sqrt2<4 Rightarrow \
0<sqrt2+sqrts_n<2 Rightarrow \
0<s_n+1<2$$
Proposition 2. $s_n$ is ascending.
We have $s_n+1=f(s_n)$ where $f(x)=sqrt2+sqrtx$. But $f'(x)=frac14 sqrt2 + sqrtx sqrtx>0$, so $f(x)$ is ascending for $x>0$. We also have that $s_2=sqrt2+sqrtsqrt2>sqrt2=s_1$, thus
$$f(s_2)geq f(s_1) iff s_3geq s_2$$
and by induction
$$f(s_n)geq f(s_n-1) iff s_n+1geq s_n$$
As a result, we have a monotone and bounded sequence, so it converges ... to the solution of
$$L=sqrt2+sqrtL$$
answered Jul 19 at 8:22
rtybase
8,86721433
add a comment |Â
up vote
0
down vote
Because nobody mentioned this method, not even in the question I flagged as duplicate ...
Proposition 1. $0<s_n <2$
By induction. $0<s_1=sqrt2<2$. Let's assume (induction hypothesis)
$$0<s_n<2 Rightarrow
0<sqrts_n<sqrt2Rightarrow \
0<2<2+sqrts_n<2+sqrt2<4 Rightarrow \
0<sqrt2+sqrts_n<2 Rightarrow \
0<s_n+1<2$$
Proposition 2. $s_n$ is ascending.
We have $s_n+1=f(s_n)$ where $f(x)=sqrt2+sqrtx$. But $f'(x)=frac14 sqrt2 + sqrtx sqrtx>0$, so $f(x)$ is ascending for $x>0$. We also have that $s_2=sqrt2+sqrtsqrt2>sqrt2=s_1$, thus
$$f(s_2)geq f(s_1) iff s_3geq s_2$$
and by induction
$$f(s_n)geq f(s_n-1) iff s_n+1geq s_n$$
As a result, we have a monotone and bounded sequence, so it converges ... to the solution of
$$L=sqrt2+sqrtL$$
answered Jul 19 at 8:22
rtybase
8,86721433
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Because nobody mentioned this method, not even in the question I flagged as duplicate ...
Proposition 1. $0<s_n <2$
By induction. $0<s_1=sqrt2<2$. Let's assume (induction hypothesis)
$$0<s_n<2 Rightarrow
0<sqrts_n<sqrt2Rightarrow \
0<2<2+sqrts_n<2+sqrt2<4 Rightarrow \
0<sqrt2+sqrts_n<2 Rightarrow \
0<s_n+1<2$$
Proposition 2. $s_n$ is ascending.
We have $s_n+1=f(s_n)$ where $f(x)=sqrt2+sqrtx$. But $f'(x)=frac14 sqrt2 + sqrtx sqrtx>0$, so $f(x)$ is ascending for $x>0$. We also have that $s_2=sqrt2+sqrtsqrt2>sqrt2=s_1$, thus
$$f(s_2)geq f(s_1) iff s_3geq s_2$$
and by induction
$$f(s_n)geq f(s_n-1) iff s_n+1geq s_n$$
As a result, we have a monotone and bounded sequence, so it converges ... to the solution of
$$L=sqrt2+sqrtL$$
answered Jul 19 at 8:22
rtybase
8,86721433
Because nobody mentioned this method, not even in the question I flagged as duplicate ...
Proposition 1. $0<s_n <2$
By induction. $0<s_1=sqrt2<2$. Let's assume (induction hypothesis)
$$0<s_n<2 Rightarrow
0<sqrts_n<sqrt2Rightarrow \
0<2<2+sqrts_n<2+sqrt2<4 Rightarrow \
0<sqrt2+sqrts_n<2 Rightarrow \
0<s_n+1<2$$
Proposition 2. $s_n$ is ascending.
We have $s_n+1=f(s_n)$ where $f(x)=sqrt2+sqrtx$. But $f'(x)=frac14 sqrt2 + sqrtx sqrtx>0$, so $f(x)$ is ascending for $x>0$. We also have that $s_2=sqrt2+sqrtsqrt2>sqrt2=s_1$, thus
$$f(s_2)geq f(s_1) iff s_3geq s_2$$
and by induction
$$f(s_n)geq f(s_n-1) iff s_n+1geq s_n$$
As a result, we have a monotone and bounded sequence, so it converges ... to the solution of
$$L=sqrt2+sqrtL$$
answered Jul 19 at 8:22
rtybase
8,86721433
edited Jul 19 at 10:31
answered Jul 19 at 8:22
rtybase
8,86721433
answered Jul 19 at 8:22
rtybase
8,86721433
answered Jul 19 at 8:22
rtybase
8,86721433
8,86721433
add a comment |Â
add a comment |Â
You can use Mathjax : math.meta.stackexchange.com/questions/5020/…
– nicomezi
Jul 19 at 7:22
How about this one math.stackexchange.com/questions/2702895/…?
– rtybase
Jul 19 at 8:00
@rtybase This one answers the second part of the question. Thanks for pointing that out!
– Miles123K
Jul 19 at 8:05