Prove that two sequences converge to the same limit.
I encountered this question in my homework:
$$a_1=x, b_1=y, \
a_{n+1} =frac{a_n+b_n}{2}, b_{n+1}= sqrt{a_nb_n}, nin mathbb{N}$$
Given $x,y$ positive constants.
I have to prove that they both converge to the same limit $L$.
I know that in order to prove that a sequence converges, it has to be (for example) Monotonically increasing and that $|a_n|<K$
I'm having difficulties proving with induction that either of the sequences is monotonic... how do I do so when the first sequence depends on the other? I know that by using the average inequality I get that $a_{n+1} > b_{n+1}$. How do I continue from here?
Thank you for your time and help!
sequences-and-series convergence induction
edited Nov 23 '18 at 13:57
Mason
1,9591530
asked Nov 23 '18 at 13:40
Buk LauBuk Lau
1878
add a comment |
I encountered this question in my homework:
$$a_1=x, b_1=y, \
a_{n+1} =frac{a_n+b_n}{2}, b_{n+1}= sqrt{a_nb_n}, nin mathbb{N}$$
Given $x,y$ positive constants.
I have to prove that they both converge to the same limit $L$.
I know that in order to prove that a sequence converges, it has to be (for example) Monotonically increasing and that $|a_n|<K$
I'm having difficulties proving with induction that either of the sequences is monotonic... how do I do so when the first sequence depends on the other? I know that by using the average inequality I get that $a_{n+1} > b_{n+1}$. How do I continue from here?
Thank you for your time and help!
sequences-and-series convergence induction
edited Nov 23 '18 at 13:57
Mason
1,9591530
asked Nov 23 '18 at 13:40
Buk LauBuk Lau
1878
$(sqrt{a_n}+sqrt{b_n})^2=a_n+b_n+2sqrt{a_nb_n} geq 0$ From here we get that $frac{a_n+b_n}{2}+sqrt{a_nb_n} geq 0$
– mathnoob
Nov 23 '18 at 14:00
add a comment |
I encountered this question in my homework:
$$a_1=x, b_1=y, \
a_{n+1} =frac{a_n+b_n}{2}, b_{n+1}= sqrt{a_nb_n}, nin mathbb{N}$$
Given $x,y$ positive constants.
I have to prove that they both converge to the same limit $L$.
I know that in order to prove that a sequence converges, it has to be (for example) Monotonically increasing and that $|a_n|<K$
I'm having difficulties proving with induction that either of the sequences is monotonic... how do I do so when the first sequence depends on the other? I know that by using the average inequality I get that $a_{n+1} > b_{n+1}$. How do I continue from here?
Thank you for your time and help!
sequences-and-series convergence induction
edited Nov 23 '18 at 13:57
Mason
1,9591530
asked Nov 23 '18 at 13:40
Buk LauBuk Lau
1878
I encountered this question in my homework:
$$a_1=x, b_1=y, \
a_{n+1} =frac{a_n+b_n}{2}, b_{n+1}= sqrt{a_nb_n}, nin mathbb{N}$$
Given $x,y$ positive constants.
I have to prove that they both converge to the same limit $L$.
I know that in order to prove that a sequence converges, it has to be (for example) Monotonically increasing and that $|a_n|<K$
I'm having difficulties proving with induction that either of the sequences is monotonic... how do I do so when the first sequence depends on the other? I know that by using the average inequality I get that $a_{n+1} > b_{n+1}$. How do I continue from here?
Thank you for your time and help!
sequences-and-series convergence induction
sequences-and-series convergence induction
edited Nov 23 '18 at 13:57
Mason
1,9591530
asked Nov 23 '18 at 13:40
Buk LauBuk Lau
1878
edited Nov 23 '18 at 13:57
Mason
1,9591530
asked Nov 23 '18 at 13:40
Buk LauBuk Lau
1878
edited Nov 23 '18 at 13:57
Mason
1,9591530
edited Nov 23 '18 at 13:57
Mason
1,9591530
edited Nov 23 '18 at 13:57
Mason
1,9591530
1,9591530
asked Nov 23 '18 at 13:40
Buk LauBuk Lau
1878
asked Nov 23 '18 at 13:40
Buk LauBuk Lau
1878
asked Nov 23 '18 at 13:40
Buk LauBuk Lau
1878
1878
$(sqrt{a_n}+sqrt{b_n})^2=a_n+b_n+2sqrt{a_nb_n} geq 0$ From here we get that $frac{a_n+b_n}{2}+sqrt{a_nb_n} geq 0$
– mathnoob
Nov 23 '18 at 14:00
add a comment |
$(sqrt{a_n}+sqrt{b_n})^2=a_n+b_n+2sqrt{a_nb_n} geq 0$ From here we get that $frac{a_n+b_n}{2}+sqrt{a_nb_n} geq 0$
– mathnoob
Nov 23 '18 at 14:00
$(sqrt{a_n}+sqrt{b_n})^2=a_n+b_n+2sqrt{a_nb_n} geq 0$ From here we get that $frac{a_n+b_n}{2}+sqrt{a_nb_n} geq 0$
– mathnoob
Nov 23 '18 at 14:00
$(sqrt{a_n}+sqrt{b_n})^2=a_n+b_n+2sqrt{a_nb_n} geq 0$ From here we get that $frac{a_n+b_n}{2}+sqrt{a_nb_n} geq 0$
– mathnoob
Nov 23 '18 at 14:00
add a comment |
3 Answers
3
active
oldest
votes
By AM-GM inequality,
$$b_{n+1} le a_{n+1}$$
Also, if $ b_n le a_n$, then $a_{n+1} le a_n$ since $a_{n+1}$ is the arithmetic means between $a_n$ and $b_n$.
Also if $b_n le a_n$, then $b_{n+1} ge b_n$ since $b_{n+1}$ is the geometric means between $a_n$ and $b_n$.
Hence we have $$b_{n+1} le b_{n+2} le a_{n+2} le a_{n+1}$$
Both sequence converges.
From
$$ a=frac{a+b}2$$
we can deduce that $a=b$.
answered Nov 23 '18 at 14:12
Siong Thye GohSiong Thye Goh
100k1465117
Does that hold for any n? Or do I have to do it through induction? Do I have to assume that x>y or the other way around? Thank you!
– Buk Lau
Nov 23 '18 at 14:16
try to prove by induction from $n ge 2$ onwards. we do not need assumption of $x>y$. it always hold for any positive pairs.
– Siong Thye Goh
Nov 23 '18 at 14:18
I never encountered such a sequence that relies on another sequencem I don't know how to even start, like how do I do it just for an for example? how do I take bn out of the picture?
– Buk Lau
Nov 23 '18 at 14:21
perhaps go slow, first check that $a_n$ and $b_n$ are positive. After that check that $a_{n+1} < b_{n+1}$. after that then prove the monotonicity. ping if you are stuck and update me which step are u stuck at?
– Siong Thye Goh
Nov 23 '18 at 14:28
imgur.com/a/TaaijH9 is this the way to do it? Or did I write nonsense? Sorry if so!! Thank you.
– Buk Lau
Nov 23 '18 at 15:01
|
show 1 more comment
if $x>y$, first observe that $x ge a_nge a_{n+1} ge b_{n+1} ge b_n ge y$ for all $ninmathbb{N}$
So both ${a_n}$ and ${b_n}$ converge. (as they are monotonic bounded)
Also $|a_n-b_n|le|frac{a_{n-1}+b_{n-1}}{2}-b_{n-1}|le frac{1}{2}|a_{n-1}-b_{n-1}|le frac{1}{2^{n-1}} |a_1-b_1| to 0 $ as $nto infty$
So ${a_n}$ and ${b_n}$ converge at same limit.
if $y>x$ then $y ge a_nge a_{n+1} ge b_{n+1} ge b_n ge x$ for all $n>1$.
answered Nov 23 '18 at 14:19
OfflawOfflaw
2649
I did get to that inequity chain but I couldn't tell if it holds for all n, I just assumed it did, since all I could say for sure is that a2>a1 (when x>y). I know that it's true but do I have to show it for all n or can I just say it the way you did? Thank you!
– Buk Lau
Nov 23 '18 at 14:40
As $x,y$ are both positive it is the standard AM-GM inequality. So you also can use the result in formal proof.
– Offlaw
Nov 23 '18 at 15:04
This is exactly what I did but I thought that I still have to prove that it does hold for all n, I then got that they converge to the same limit the same way Siong Thye Goh mentioned in his answer. will this be enough as a formal proof? Thank you for your time!
– Buk Lau
Nov 23 '18 at 15:16
add a comment |
$|a_{n+1}-b_{n+1}|=left|frac{a_n+b_n}{2}-sqrt{a_nb_n}right|=1/2left|a_n+b_n-2sqrt{a_nb_n}right|=1/2(sqrt{a_n}-sqrt{b_n})^2=(1/2)(1/4)(sqrt{a_{n-1}}-sqrt{b_{n-1}})^4=2^{-1-2}(sqrt{a_{n-1}}-sqrt{b_{n-1}})^{2times 2}=ldots=2^{-1-2-ldots-n}(sqrt{x}-sqrt{y})^{2n}=2^{-n(n+1)/2}(sqrt{x}-sqrt{y})^{2n}rightarrow 0$ as $nrightarrowinfty.$
answered Nov 23 '18 at 14:00
John_WickJohn_Wick
1,486111
How did you get the third equality?
– mathnoob
Nov 23 '18 at 14:09
add a comment |
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3 Answers
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active
oldest
votes
3 Answers
3
active
oldest
votes
active
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oldest
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By AM-GM inequality,
$$b_{n+1} le a_{n+1}$$
Also, if $ b_n le a_n$, then $a_{n+1} le a_n$ since $a_{n+1}$ is the arithmetic means between $a_n$ and $b_n$.
Also if $b_n le a_n$, then $b_{n+1} ge b_n$ since $b_{n+1}$ is the geometric means between $a_n$ and $b_n$.
Hence we have $$b_{n+1} le b_{n+2} le a_{n+2} le a_{n+1}$$
Both sequence converges.
From
$$ a=frac{a+b}2$$
we can deduce that $a=b$.
answered Nov 23 '18 at 14:12
Siong Thye GohSiong Thye Goh
100k1465117
Does that hold for any n? Or do I have to do it through induction? Do I have to assume that x>y or the other way around? Thank you!
– Buk Lau
Nov 23 '18 at 14:16
try to prove by induction from $n ge 2$ onwards. we do not need assumption of $x>y$. it always hold for any positive pairs.
– Siong Thye Goh
Nov 23 '18 at 14:18
I never encountered such a sequence that relies on another sequencem I don't know how to even start, like how do I do it just for an for example? how do I take bn out of the picture?
– Buk Lau
Nov 23 '18 at 14:21
perhaps go slow, first check that $a_n$ and $b_n$ are positive. After that check that $a_{n+1} < b_{n+1}$. after that then prove the monotonicity. ping if you are stuck and update me which step are u stuck at?
– Siong Thye Goh
Nov 23 '18 at 14:28
imgur.com/a/TaaijH9 is this the way to do it? Or did I write nonsense? Sorry if so!! Thank you.
– Buk Lau
Nov 23 '18 at 15:01
|
show 1 more comment
By AM-GM inequality,
$$b_{n+1} le a_{n+1}$$
Also, if $ b_n le a_n$, then $a_{n+1} le a_n$ since $a_{n+1}$ is the arithmetic means between $a_n$ and $b_n$.
Also if $b_n le a_n$, then $b_{n+1} ge b_n$ since $b_{n+1}$ is the geometric means between $a_n$ and $b_n$.
Hence we have $$b_{n+1} le b_{n+2} le a_{n+2} le a_{n+1}$$
Both sequence converges.
From
$$ a=frac{a+b}2$$
we can deduce that $a=b$.
answered Nov 23 '18 at 14:12
Siong Thye GohSiong Thye Goh
100k1465117
Does that hold for any n? Or do I have to do it through induction? Do I have to assume that x>y or the other way around? Thank you!
– Buk Lau
Nov 23 '18 at 14:16
try to prove by induction from $n ge 2$ onwards. we do not need assumption of $x>y$. it always hold for any positive pairs.
– Siong Thye Goh
Nov 23 '18 at 14:18
I never encountered such a sequence that relies on another sequencem I don't know how to even start, like how do I do it just for an for example? how do I take bn out of the picture?
– Buk Lau
Nov 23 '18 at 14:21
perhaps go slow, first check that $a_n$ and $b_n$ are positive. After that check that $a_{n+1} < b_{n+1}$. after that then prove the monotonicity. ping if you are stuck and update me which step are u stuck at?
– Siong Thye Goh
Nov 23 '18 at 14:28
imgur.com/a/TaaijH9 is this the way to do it? Or did I write nonsense? Sorry if so!! Thank you.
– Buk Lau
Nov 23 '18 at 15:01
|
show 1 more comment
By AM-GM inequality,
$$b_{n+1} le a_{n+1}$$
Also, if $ b_n le a_n$, then $a_{n+1} le a_n$ since $a_{n+1}$ is the arithmetic means between $a_n$ and $b_n$.
Also if $b_n le a_n$, then $b_{n+1} ge b_n$ since $b_{n+1}$ is the geometric means between $a_n$ and $b_n$.
Hence we have $$b_{n+1} le b_{n+2} le a_{n+2} le a_{n+1}$$
Both sequence converges.
From
$$ a=frac{a+b}2$$
we can deduce that $a=b$.
answered Nov 23 '18 at 14:12
Siong Thye GohSiong Thye Goh
100k1465117
By AM-GM inequality,
$$b_{n+1} le a_{n+1}$$
Also, if $ b_n le a_n$, then $a_{n+1} le a_n$ since $a_{n+1}$ is the arithmetic means between $a_n$ and $b_n$.
Also if $b_n le a_n$, then $b_{n+1} ge b_n$ since $b_{n+1}$ is the geometric means between $a_n$ and $b_n$.
Hence we have $$b_{n+1} le b_{n+2} le a_{n+2} le a_{n+1}$$
Both sequence converges.
From
$$ a=frac{a+b}2$$
we can deduce that $a=b$.
answered Nov 23 '18 at 14:12
Siong Thye GohSiong Thye Goh
100k1465117
answered Nov 23 '18 at 14:12
Siong Thye GohSiong Thye Goh
100k1465117
answered Nov 23 '18 at 14:12
Siong Thye GohSiong Thye Goh
100k1465117
answered Nov 23 '18 at 14:12
Siong Thye GohSiong Thye Goh
100k1465117
100k1465117
Does that hold for any n? Or do I have to do it through induction? Do I have to assume that x>y or the other way around? Thank you!
– Buk Lau
Nov 23 '18 at 14:16
try to prove by induction from $n ge 2$ onwards. we do not need assumption of $x>y$. it always hold for any positive pairs.
– Siong Thye Goh
Nov 23 '18 at 14:18
I never encountered such a sequence that relies on another sequencem I don't know how to even start, like how do I do it just for an for example? how do I take bn out of the picture?
– Buk Lau
Nov 23 '18 at 14:21
perhaps go slow, first check that $a_n$ and $b_n$ are positive. After that check that $a_{n+1} < b_{n+1}$. after that then prove the monotonicity. ping if you are stuck and update me which step are u stuck at?
– Siong Thye Goh
Nov 23 '18 at 14:28
imgur.com/a/TaaijH9 is this the way to do it? Or did I write nonsense? Sorry if so!! Thank you.
– Buk Lau
Nov 23 '18 at 15:01
|
show 1 more comment
Does that hold for any n? Or do I have to do it through induction? Do I have to assume that x>y or the other way around? Thank you!
– Buk Lau
Nov 23 '18 at 14:16
try to prove by induction from $n ge 2$ onwards. we do not need assumption of $x>y$. it always hold for any positive pairs.
– Siong Thye Goh
Nov 23 '18 at 14:18
I never encountered such a sequence that relies on another sequencem I don't know how to even start, like how do I do it just for an for example? how do I take bn out of the picture?
– Buk Lau
Nov 23 '18 at 14:21
perhaps go slow, first check that $a_n$ and $b_n$ are positive. After that check that $a_{n+1} < b_{n+1}$. after that then prove the monotonicity. ping if you are stuck and update me which step are u stuck at?
– Siong Thye Goh
Nov 23 '18 at 14:28
imgur.com/a/TaaijH9 is this the way to do it? Or did I write nonsense? Sorry if so!! Thank you.
– Buk Lau
Nov 23 '18 at 15:01
Does that hold for any n? Or do I have to do it through induction? Do I have to assume that x>y or the other way around? Thank you!
– Buk Lau
Nov 23 '18 at 14:16
Does that hold for any n? Or do I have to do it through induction? Do I have to assume that x>y or the other way around? Thank you!
– Buk Lau
Nov 23 '18 at 14:16
try to prove by induction from $n ge 2$ onwards. we do not need assumption of $x>y$. it always hold for any positive pairs.
– Siong Thye Goh
Nov 23 '18 at 14:18
try to prove by induction from $n ge 2$ onwards. we do not need assumption of $x>y$. it always hold for any positive pairs.
– Siong Thye Goh
Nov 23 '18 at 14:18
I never encountered such a sequence that relies on another sequencem I don't know how to even start, like how do I do it just for an for example? how do I take bn out of the picture?
– Buk Lau
Nov 23 '18 at 14:21
I never encountered such a sequence that relies on another sequencem I don't know how to even start, like how do I do it just for an for example? how do I take bn out of the picture?
– Buk Lau
Nov 23 '18 at 14:21
perhaps go slow, first check that $a_n$ and $b_n$ are positive. After that check that $a_{n+1} < b_{n+1}$. after that then prove the monotonicity. ping if you are stuck and update me which step are u stuck at?
– Siong Thye Goh
Nov 23 '18 at 14:28
perhaps go slow, first check that $a_n$ and $b_n$ are positive. After that check that $a_{n+1} < b_{n+1}$. after that then prove the monotonicity. ping if you are stuck and update me which step are u stuck at?
– Siong Thye Goh
Nov 23 '18 at 14:28
imgur.com/a/TaaijH9 is this the way to do it? Or did I write nonsense? Sorry if so!! Thank you.
– Buk Lau
Nov 23 '18 at 15:01
imgur.com/a/TaaijH9 is this the way to do it? Or did I write nonsense? Sorry if so!! Thank you.
– Buk Lau
Nov 23 '18 at 15:01
|
show 1 more comment
if $x>y$, first observe that $x ge a_nge a_{n+1} ge b_{n+1} ge b_n ge y$ for all $ninmathbb{N}$
So both ${a_n}$ and ${b_n}$ converge. (as they are monotonic bounded)
Also $|a_n-b_n|le|frac{a_{n-1}+b_{n-1}}{2}-b_{n-1}|le frac{1}{2}|a_{n-1}-b_{n-1}|le frac{1}{2^{n-1}} |a_1-b_1| to 0 $ as $nto infty$
So ${a_n}$ and ${b_n}$ converge at same limit.
if $y>x$ then $y ge a_nge a_{n+1} ge b_{n+1} ge b_n ge x$ for all $n>1$.
answered Nov 23 '18 at 14:19
OfflawOfflaw
2649
I did get to that inequity chain but I couldn't tell if it holds for all n, I just assumed it did, since all I could say for sure is that a2>a1 (when x>y). I know that it's true but do I have to show it for all n or can I just say it the way you did? Thank you!
– Buk Lau
Nov 23 '18 at 14:40
As $x,y$ are both positive it is the standard AM-GM inequality. So you also can use the result in formal proof.
– Offlaw
Nov 23 '18 at 15:04
This is exactly what I did but I thought that I still have to prove that it does hold for all n, I then got that they converge to the same limit the same way Siong Thye Goh mentioned in his answer. will this be enough as a formal proof? Thank you for your time!
– Buk Lau
Nov 23 '18 at 15:16
add a comment |
if $x>y$, first observe that $x ge a_nge a_{n+1} ge b_{n+1} ge b_n ge y$ for all $ninmathbb{N}$
So both ${a_n}$ and ${b_n}$ converge. (as they are monotonic bounded)
Also $|a_n-b_n|le|frac{a_{n-1}+b_{n-1}}{2}-b_{n-1}|le frac{1}{2}|a_{n-1}-b_{n-1}|le frac{1}{2^{n-1}} |a_1-b_1| to 0 $ as $nto infty$
So ${a_n}$ and ${b_n}$ converge at same limit.
if $y>x$ then $y ge a_nge a_{n+1} ge b_{n+1} ge b_n ge x$ for all $n>1$.
answered Nov 23 '18 at 14:19
OfflawOfflaw
2649
I did get to that inequity chain but I couldn't tell if it holds for all n, I just assumed it did, since all I could say for sure is that a2>a1 (when x>y). I know that it's true but do I have to show it for all n or can I just say it the way you did? Thank you!
– Buk Lau
Nov 23 '18 at 14:40
As $x,y$ are both positive it is the standard AM-GM inequality. So you also can use the result in formal proof.
– Offlaw
Nov 23 '18 at 15:04
This is exactly what I did but I thought that I still have to prove that it does hold for all n, I then got that they converge to the same limit the same way Siong Thye Goh mentioned in his answer. will this be enough as a formal proof? Thank you for your time!
– Buk Lau
Nov 23 '18 at 15:16
add a comment |
if $x>y$, first observe that $x ge a_nge a_{n+1} ge b_{n+1} ge b_n ge y$ for all $ninmathbb{N}$
So both ${a_n}$ and ${b_n}$ converge. (as they are monotonic bounded)
Also $|a_n-b_n|le|frac{a_{n-1}+b_{n-1}}{2}-b_{n-1}|le frac{1}{2}|a_{n-1}-b_{n-1}|le frac{1}{2^{n-1}} |a_1-b_1| to 0 $ as $nto infty$
So ${a_n}$ and ${b_n}$ converge at same limit.
if $y>x$ then $y ge a_nge a_{n+1} ge b_{n+1} ge b_n ge x$ for all $n>1$.
answered Nov 23 '18 at 14:19
OfflawOfflaw
2649
if $x>y$, first observe that $x ge a_nge a_{n+1} ge b_{n+1} ge b_n ge y$ for all $ninmathbb{N}$
So both ${a_n}$ and ${b_n}$ converge. (as they are monotonic bounded)
Also $|a_n-b_n|le|frac{a_{n-1}+b_{n-1}}{2}-b_{n-1}|le frac{1}{2}|a_{n-1}-b_{n-1}|le frac{1}{2^{n-1}} |a_1-b_1| to 0 $ as $nto infty$
So ${a_n}$ and ${b_n}$ converge at same limit.
if $y>x$ then $y ge a_nge a_{n+1} ge b_{n+1} ge b_n ge x$ for all $n>1$.
answered Nov 23 '18 at 14:19
OfflawOfflaw
2649
answered Nov 23 '18 at 14:19
OfflawOfflaw
2649
answered Nov 23 '18 at 14:19
OfflawOfflaw
2649
answered Nov 23 '18 at 14:19
OfflawOfflaw
2649
2649
I did get to that inequity chain but I couldn't tell if it holds for all n, I just assumed it did, since all I could say for sure is that a2>a1 (when x>y). I know that it's true but do I have to show it for all n or can I just say it the way you did? Thank you!
– Buk Lau
Nov 23 '18 at 14:40
As $x,y$ are both positive it is the standard AM-GM inequality. So you also can use the result in formal proof.
– Offlaw
Nov 23 '18 at 15:04
This is exactly what I did but I thought that I still have to prove that it does hold for all n, I then got that they converge to the same limit the same way Siong Thye Goh mentioned in his answer. will this be enough as a formal proof? Thank you for your time!
– Buk Lau
Nov 23 '18 at 15:16
add a comment |
I did get to that inequity chain but I couldn't tell if it holds for all n, I just assumed it did, since all I could say for sure is that a2>a1 (when x>y). I know that it's true but do I have to show it for all n or can I just say it the way you did? Thank you!
– Buk Lau
Nov 23 '18 at 14:40
As $x,y$ are both positive it is the standard AM-GM inequality. So you also can use the result in formal proof.
– Offlaw
Nov 23 '18 at 15:04
This is exactly what I did but I thought that I still have to prove that it does hold for all n, I then got that they converge to the same limit the same way Siong Thye Goh mentioned in his answer. will this be enough as a formal proof? Thank you for your time!
– Buk Lau
Nov 23 '18 at 15:16
I did get to that inequity chain but I couldn't tell if it holds for all n, I just assumed it did, since all I could say for sure is that a2>a1 (when x>y). I know that it's true but do I have to show it for all n or can I just say it the way you did? Thank you!
– Buk Lau
Nov 23 '18 at 14:40
I did get to that inequity chain but I couldn't tell if it holds for all n, I just assumed it did, since all I could say for sure is that a2>a1 (when x>y). I know that it's true but do I have to show it for all n or can I just say it the way you did? Thank you!
– Buk Lau
Nov 23 '18 at 14:40
As $x,y$ are both positive it is the standard AM-GM inequality. So you also can use the result in formal proof.
– Offlaw
Nov 23 '18 at 15:04
As $x,y$ are both positive it is the standard AM-GM inequality. So you also can use the result in formal proof.
– Offlaw
Nov 23 '18 at 15:04
This is exactly what I did but I thought that I still have to prove that it does hold for all n, I then got that they converge to the same limit the same way Siong Thye Goh mentioned in his answer. will this be enough as a formal proof? Thank you for your time!
– Buk Lau
Nov 23 '18 at 15:16
This is exactly what I did but I thought that I still have to prove that it does hold for all n, I then got that they converge to the same limit the same way Siong Thye Goh mentioned in his answer. will this be enough as a formal proof? Thank you for your time!
– Buk Lau
Nov 23 '18 at 15:16
add a comment |
$|a_{n+1}-b_{n+1}|=left|frac{a_n+b_n}{2}-sqrt{a_nb_n}right|=1/2left|a_n+b_n-2sqrt{a_nb_n}right|=1/2(sqrt{a_n}-sqrt{b_n})^2=(1/2)(1/4)(sqrt{a_{n-1}}-sqrt{b_{n-1}})^4=2^{-1-2}(sqrt{a_{n-1}}-sqrt{b_{n-1}})^{2times 2}=ldots=2^{-1-2-ldots-n}(sqrt{x}-sqrt{y})^{2n}=2^{-n(n+1)/2}(sqrt{x}-sqrt{y})^{2n}rightarrow 0$ as $nrightarrowinfty.$
answered Nov 23 '18 at 14:00
John_WickJohn_Wick
1,486111
How did you get the third equality?
– mathnoob
Nov 23 '18 at 14:09
add a comment |
$|a_{n+1}-b_{n+1}|=left|frac{a_n+b_n}{2}-sqrt{a_nb_n}right|=1/2left|a_n+b_n-2sqrt{a_nb_n}right|=1/2(sqrt{a_n}-sqrt{b_n})^2=(1/2)(1/4)(sqrt{a_{n-1}}-sqrt{b_{n-1}})^4=2^{-1-2}(sqrt{a_{n-1}}-sqrt{b_{n-1}})^{2times 2}=ldots=2^{-1-2-ldots-n}(sqrt{x}-sqrt{y})^{2n}=2^{-n(n+1)/2}(sqrt{x}-sqrt{y})^{2n}rightarrow 0$ as $nrightarrowinfty.$
answered Nov 23 '18 at 14:00
John_WickJohn_Wick
1,486111
How did you get the third equality?
– mathnoob
Nov 23 '18 at 14:09
add a comment |
$|a_{n+1}-b_{n+1}|=left|frac{a_n+b_n}{2}-sqrt{a_nb_n}right|=1/2left|a_n+b_n-2sqrt{a_nb_n}right|=1/2(sqrt{a_n}-sqrt{b_n})^2=(1/2)(1/4)(sqrt{a_{n-1}}-sqrt{b_{n-1}})^4=2^{-1-2}(sqrt{a_{n-1}}-sqrt{b_{n-1}})^{2times 2}=ldots=2^{-1-2-ldots-n}(sqrt{x}-sqrt{y})^{2n}=2^{-n(n+1)/2}(sqrt{x}-sqrt{y})^{2n}rightarrow 0$ as $nrightarrowinfty.$
answered Nov 23 '18 at 14:00
John_WickJohn_Wick
1,486111
$|a_{n+1}-b_{n+1}|=left|frac{a_n+b_n}{2}-sqrt{a_nb_n}right|=1/2left|a_n+b_n-2sqrt{a_nb_n}right|=1/2(sqrt{a_n}-sqrt{b_n})^2=(1/2)(1/4)(sqrt{a_{n-1}}-sqrt{b_{n-1}})^4=2^{-1-2}(sqrt{a_{n-1}}-sqrt{b_{n-1}})^{2times 2}=ldots=2^{-1-2-ldots-n}(sqrt{x}-sqrt{y})^{2n}=2^{-n(n+1)/2}(sqrt{x}-sqrt{y})^{2n}rightarrow 0$ as $nrightarrowinfty.$
answered Nov 23 '18 at 14:00
John_WickJohn_Wick
1,486111
edited Nov 23 '18 at 14:13
answered Nov 23 '18 at 14:00
John_WickJohn_Wick
1,486111
answered Nov 23 '18 at 14:00
John_WickJohn_Wick
1,486111
answered Nov 23 '18 at 14:00
John_WickJohn_Wick
1,486111
1,486111
How did you get the third equality?
– mathnoob
Nov 23 '18 at 14:09
add a comment |
How did you get the third equality?
– mathnoob
Nov 23 '18 at 14:09
How did you get the third equality?
– mathnoob
Nov 23 '18 at 14:09
How did you get the third equality?
– mathnoob
Nov 23 '18 at 14:09
add a comment |
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$(sqrt{a_n}+sqrt{b_n})^2=a_n+b_n+2sqrt{a_nb_n} geq 0$ From here we get that $frac{a_n+b_n}{2}+sqrt{a_nb_n} geq 0$
– mathnoob
Nov 23 '18 at 14:00