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Arithmetic-Geometric mean sequences limit [duplicate]

Show that $lim_ntoinftyb_n=fracsqrtb^2-a^2arccosfracab$Limit of product series for convergent and increasing sequencesWorking out values of sequences using partial sumsLimit of ratio of sequencesEvaluate $lim_nrightarrowinfty fraca_nb_n$Show that there are rectangles $R_1,…,R_m$ such that $F = cup_k=1^mR_k$ and diam$R_k < epsilon$ for each $k$.Show that $lim_ntoinftyb_n=fracsqrtb^2-a^2arccosfracab$Sequence and series. What is $b_n$?Find all possible sequences $a_n$ and $b_n$ such that $f$ is continuous on $[0,infty)$.Finding $sum_i=1^100 a_i$ given that $sqrta_1+sqrta_2-1+sqrta_3-2+dots+sqrta_n-(n-1)=frac12(a_1+a_2+dots+a_n)=fracn(n-3)4$Proving convergence and finding limit of two sequences defined indutively

2

1

$begingroup$

This question already has an answer here:

• 
  Show that $lim_ntoinftyb_n=fracsqrtb^2-a^2arccosfracab$
  
  2 answers
  
  

If $a,b$ are positive quantities such that $(a<b)$ and if

beginalign
a_1 &= fraca+b2 & b_1 & = sqrta_1 b \
a_2 &= fraca_1+b_12 & b_2 &= sqrta_2 b_1 \
&phantom36pt vdots & & \
a_n &= fraca_n-1+b_n-12 & b_n & = sqrta_n b_n-1 \
&phantom36pt vdots & &
endalign then show that $displaystylelim_limitsnto infty b_n=fracsqrtb^2-a^2arccos(fracab)$

real-analysis sequences-and-series limits

share|cite|improve this question

edited Mar 15 at 1:47

Lee David Chung Lin

4,39341242

asked Jun 22 '17 at 20:54

Yuriy SolovyovYuriy Solovyov

955

$endgroup$

marked as duplicate by Arnaud D., Lee David Chung Lin, Song, YiFan, Lord Shark the Unknown Mar 15 at 5:35

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 | 

2

1

$begingroup$

This question already has an answer here:

• 
  Show that $lim_ntoinftyb_n=fracsqrtb^2-a^2arccosfracab$
  
  2 answers
  
  

If $a,b$ are positive quantities such that $(a<b)$ and if

beginalign
a_1 &= fraca+b2 & b_1 & = sqrta_1 b \
a_2 &= fraca_1+b_12 & b_2 &= sqrta_2 b_1 \
&phantom36pt vdots & & \
a_n &= fraca_n-1+b_n-12 & b_n & = sqrta_n b_n-1 \
&phantom36pt vdots & &
endalign then show that $displaystylelim_limitsnto infty b_n=fracsqrtb^2-a^2arccos(fracab)$

real-analysis sequences-and-series limits

share|cite|improve this question

edited Mar 15 at 1:47

Lee David Chung Lin

4,39341242

asked Jun 22 '17 at 20:54

Yuriy SolovyovYuriy Solovyov

955

$endgroup$

marked as duplicate by Arnaud D., Lee David Chung Lin, Song, YiFan, Lord Shark the Unknown Mar 15 at 5:35

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 | 

$begingroup$

This question already has an answer here:

• 
  Show that $lim_ntoinftyb_n=fracsqrtb^2-a^2arccosfracab$
  
  2 answers
  
  

If $a,b$ are positive quantities such that $(a<b)$ and if

beginalign
a_1 &= fraca+b2 & b_1 & = sqrta_1 b \
a_2 &= fraca_1+b_12 & b_2 &= sqrta_2 b_1 \
&phantom36pt vdots & & \
a_n &= fraca_n-1+b_n-12 & b_n & = sqrta_n b_n-1 \
&phantom36pt vdots & &
endalign then show that $displaystylelim_limitsnto infty b_n=fracsqrtb^2-a^2arccos(fracab)$

real-analysis sequences-and-series limits

share|cite|improve this question

edited Mar 15 at 1:47

Lee David Chung Lin

4,39341242

asked Jun 22 '17 at 20:54

Yuriy SolovyovYuriy Solovyov

955

$endgroup$

share|cite|improve this question

edited Mar 15 at 1:47

Lee David Chung Lin

4,39341242

asked Jun 22 '17 at 20:54

Yuriy SolovyovYuriy Solovyov

955

share|cite|improve this question

edited Mar 15 at 1:47

Lee David Chung Lin

4,39341242

asked Jun 22 '17 at 20:54

Yuriy SolovyovYuriy Solovyov

955

edited Mar 15 at 1:47

Lee David Chung Lin

4,39341242

edited Mar 15 at 1:47

Lee David Chung Lin

4,39341242

asked Jun 22 '17 at 20:54

Yuriy SolovyovYuriy Solovyov

955

asked Jun 22 '17 at 20:54

Yuriy SolovyovYuriy Solovyov

955

1 Answer
1

active

oldest

votes

2

$begingroup$

If we assume that $b=1$ and $a=costheta$ with $thetainleft(0,fracpi2right)$ we get:

$$ a_1 = frac1+costheta2 = cos^2fractheta2,qquad b_1=cosfractheta2$$

$$ a_2 = cosfractheta2left(frac1+cosfractheta22right),qquad b_2=cosfractheta2cosfractheta4$$
so by induction it follows that
$$ b_n = prod_k=1^ncosfractheta2^k = fracsintheta2^nsinfractheta2^n $$
and $lim_nto +inftyb_n = fracsinthetatheta. $ The claim easily follows by rescaling, since $theta=arccosfracab$.

share|cite|improve this answer

edited Jun 22 '17 at 21:37

answered Jun 22 '17 at 21:10

Jack D'AurizioJack D'Aurizio

291k33284669

$endgroup$

add a comment | 

2

$begingroup$

If we assume that $b=1$ and $a=costheta$ with $thetainleft(0,fracpi2right)$ we get:

$$ a_1 = frac1+costheta2 = cos^2fractheta2,qquad b_1=cosfractheta2$$

$$ a_2 = cosfractheta2left(frac1+cosfractheta22right),qquad b_2=cosfractheta2cosfractheta4$$
so by induction it follows that
$$ b_n = prod_k=1^ncosfractheta2^k = fracsintheta2^nsinfractheta2^n $$
and $lim_nto +inftyb_n = fracsinthetatheta. $ The claim easily follows by rescaling, since $theta=arccosfracab$.

share|cite|improve this answer

edited Jun 22 '17 at 21:37

answered Jun 22 '17 at 21:10

Jack D'AurizioJack D'Aurizio

291k33284669

$endgroup$

add a comment | 

2

$begingroup$

If we assume that $b=1$ and $a=costheta$ with $thetainleft(0,fracpi2right)$ we get:

$$ a_1 = frac1+costheta2 = cos^2fractheta2,qquad b_1=cosfractheta2$$

$$ a_2 = cosfractheta2left(frac1+cosfractheta22right),qquad b_2=cosfractheta2cosfractheta4$$
so by induction it follows that
$$ b_n = prod_k=1^ncosfractheta2^k = fracsintheta2^nsinfractheta2^n $$
and $lim_nto +inftyb_n = fracsinthetatheta. $ The claim easily follows by rescaling, since $theta=arccosfracab$.

share|cite|improve this answer

edited Jun 22 '17 at 21:37

answered Jun 22 '17 at 21:10

Jack D'AurizioJack D'Aurizio

291k33284669

$endgroup$

add a comment | 

$begingroup$

If we assume that $b=1$ and $a=costheta$ with $thetainleft(0,fracpi2right)$ we get:

$$ a_1 = frac1+costheta2 = cos^2fractheta2,qquad b_1=cosfractheta2$$

$$ a_2 = cosfractheta2left(frac1+cosfractheta22right),qquad b_2=cosfractheta2cosfractheta4$$
so by induction it follows that
$$ b_n = prod_k=1^ncosfractheta2^k = fracsintheta2^nsinfractheta2^n $$
and $lim_nto +inftyb_n = fracsinthetatheta. $ The claim easily follows by rescaling, since $theta=arccosfracab$.

share|cite|improve this answer

edited Jun 22 '17 at 21:37

answered Jun 22 '17 at 21:10

Jack D'AurizioJack D'Aurizio

291k33284669

$endgroup$

share|cite|improve this answer

edited Jun 22 '17 at 21:37

answered Jun 22 '17 at 21:10

Jack D'AurizioJack D'Aurizio

291k33284669

answered Jun 22 '17 at 21:10

Jack D'AurizioJack D'Aurizio

291k33284669

answered Jun 22 '17 at 21:10

Jack D'AurizioJack D'Aurizio

291k33284669