Prove the sequence $a_n =frac{1cdot 3cdot…cdot(2n-1)}{2cdot4cdot…cdot2n}$ has a limit [duplicate]
This question already has an answer here:
What is the limit of $frac{prod Odd}{prod Even}?!$
7 answers
I have several questions to ask:
1) Show increasing, find the upper bound if you can of
$sqrt{(n^2-1)}/n$.
$sqrt{(n^2-1)}/n= |n|sqrt{1-1/n^2}/n$ if $n$ is positive than $sqrt1$ else $-sqrt1$;
bound: $sqrt{(n^2-1)}/n le sqrt {n^2}/n = |n|/n=1$
To show if it is increasing should I do $frac{sqrt{(n+1)^2-1}}{n+1} ge frac{sqrt{(n)^2-1}}{n}$?
2)Prove the sequence $a_n = frac{1cdot 3cdot…cdot(2n-1)}{2cdot4cdot…cdot2n}$ has a limit
$a_n$ is a decreasing sequence, so to have a limit it must be bounded from below.
$a_n = (1-1/2)(1-1/4)…(1-1/2n)$
real-analysis sequences-and-series limits analysis
marked as duplicate by Nosrati, Jyrki Lahtonen, Mostafa Ayaz, Did
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Nov 25 at 9:25
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 |
This question already has an answer here:
What is the limit of $frac{prod Odd}{prod Even}?!$
7 answers
I have several questions to ask:
1) Show increasing, find the upper bound if you can of
$sqrt{(n^2-1)}/n$.
$sqrt{(n^2-1)}/n= |n|sqrt{1-1/n^2}/n$ if $n$ is positive than $sqrt1$ else $-sqrt1$;
bound: $sqrt{(n^2-1)}/n le sqrt {n^2}/n = |n|/n=1$
To show if it is increasing should I do $frac{sqrt{(n+1)^2-1}}{n+1} ge frac{sqrt{(n)^2-1}}{n}$?
2)Prove the sequence $a_n = frac{1cdot 3cdot…cdot(2n-1)}{2cdot4cdot…cdot2n}$ has a limit
$a_n$ is a decreasing sequence, so to have a limit it must be bounded from below.
$a_n = (1-1/2)(1-1/4)…(1-1/2n)$
real-analysis sequences-and-series limits analysis
marked as duplicate by Nosrati, Jyrki Lahtonen, Mostafa Ayaz, Did
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Nov 25 at 9:25
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.
1
If you have several questions to ask, it is better to ask separate questions, unless the questions are closely related.
– robjohn♦
Nov 25 at 1:34
add a comment |
This question already has an answer here:
What is the limit of $frac{prod Odd}{prod Even}?!$
7 answers
I have several questions to ask:
1) Show increasing, find the upper bound if you can of
$sqrt{(n^2-1)}/n$.
$sqrt{(n^2-1)}/n= |n|sqrt{1-1/n^2}/n$ if $n$ is positive than $sqrt1$ else $-sqrt1$;
bound: $sqrt{(n^2-1)}/n le sqrt {n^2}/n = |n|/n=1$
To show if it is increasing should I do $frac{sqrt{(n+1)^2-1}}{n+1} ge frac{sqrt{(n)^2-1}}{n}$?
2)Prove the sequence $a_n = frac{1cdot 3cdot…cdot(2n-1)}{2cdot4cdot…cdot2n}$ has a limit
$a_n$ is a decreasing sequence, so to have a limit it must be bounded from below.
$a_n = (1-1/2)(1-1/4)…(1-1/2n)$
real-analysis sequences-and-series limits analysis
This question already has an answer here:
What is the limit of $frac{prod Odd}{prod Even}?!$
7 answers
I have several questions to ask:
1) Show increasing, find the upper bound if you can of
$sqrt{(n^2-1)}/n$.
$sqrt{(n^2-1)}/n= |n|sqrt{1-1/n^2}/n$ if $n$ is positive than $sqrt1$ else $-sqrt1$;
bound: $sqrt{(n^2-1)}/n le sqrt {n^2}/n = |n|/n=1$
To show if it is increasing should I do $frac{sqrt{(n+1)^2-1}}{n+1} ge frac{sqrt{(n)^2-1}}{n}$?
2)Prove the sequence $a_n = frac{1cdot 3cdot…cdot(2n-1)}{2cdot4cdot…cdot2n}$ has a limit
$a_n$ is a decreasing sequence, so to have a limit it must be bounded from below.
$a_n = (1-1/2)(1-1/4)…(1-1/2n)$
This question already has an answer here:
What is the limit of $frac{prod Odd}{prod Even}?!$
7 answers
real-analysis sequences-and-series limits analysis
real-analysis sequences-and-series limits analysis
edited Nov 24 at 22:40
asked Nov 24 at 22:06
Sargis Iskandaryan
560112
560112
marked as duplicate by Nosrati, Jyrki Lahtonen, Mostafa Ayaz, Did
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Nov 25 at 9:25
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 Nosrati, Jyrki Lahtonen, Mostafa Ayaz, Did
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Nov 25 at 9:25
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.
1
If you have several questions to ask, it is better to ask separate questions, unless the questions are closely related.
– robjohn♦
Nov 25 at 1:34
add a comment |
1
If you have several questions to ask, it is better to ask separate questions, unless the questions are closely related.
– robjohn♦
Nov 25 at 1:34
1
1
If you have several questions to ask, it is better to ask separate questions, unless the questions are closely related.
– robjohn♦
Nov 25 at 1:34
If you have several questions to ask, it is better to ask separate questions, unless the questions are closely related.
– robjohn♦
Nov 25 at 1:34
add a comment |
4 Answers
4
active
oldest
votes
Cross-multiplication yields, for $kge1$,
$$
left(frac{2k-1}{2k}right)^2lefrac{2k-1}{2k+1}
$$
Therefore,
$$
begin{align}
prod_{k=1}^nleft(frac{2k-1}{2k}right)^2
&leprod_{k=1}^nfrac{2k-1}{2k+1}\
&=frac1{2n+1}
end{align}
$$
Thus,
$$
prod_{k=1}^inftyfrac{2k-1}{2k}=0
$$
For $nge1$,
$$
begin{align}
frac{frac{sqrt{n^2-1}}n}{frac{sqrt{(n+1)^2-1}}{n+1}}
&=sqrt{frac{(n+1)^3(n-1)}{n^3(n+2)}}\
&=sqrt{frac{n^4+2n^3-2n-1}{n^4+2n^3}}\[18pt]
<1
end{align}
$$
Thus, $frac{sqrt{n^2-1}}n$ is increasing. Furthermore,
$$
begin{align}
lim_{ntoinfty}frac{sqrt{n^2-1}}n
&=lim_{ntoinfty}sqrt{1-frac1{n^2}}\[6pt]
&=1
end{align}
$$
add a comment |
HINT
We have that
$$a_n=frac{1cdot 3cdot…cdot(2n-1)}{2cdot4cdot…cdot2n}=frac{(2n)!}{(2^nn!)^2}$$
then we can use Stirling's approximation.
Refer to the related
- What is the limit of $frac{prod Odd}{prod Even}?!$
Can you help with the other two?
– Sargis Iskandaryan
Nov 24 at 22:22
@SargisIskandaryan The second one is trivial indeed $frac{n^n}{n!}=frac{ncdot nldots n}{1cdot 2cdot n}$ or by ratio test for example.
– gimusi
Nov 24 at 22:28
@SargisIskandaryan I don't understand the first question.
– gimusi
Nov 24 at 22:29
add a comment |
$$frac{(2n-1)!!}{(2n)!!}=frac{1}{4^n}binom{2n}{n}=frac{2}{pi}int_{0}^{pi/2}left(costhetaright)^{2n},dtheta $$
is quite clearly decreasing and convergent to zero (by the dominated convergence theorem).
Laplace's method gives $frac{1}{4^n}binom{2n}{n}simfrac{1}{sqrt{pileft(n+frac{1}{4}right)}}$.
add a comment |
The sequence is decreasing and bounded below by 0. That's why it has a limit.
add a comment |
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
Cross-multiplication yields, for $kge1$,
$$
left(frac{2k-1}{2k}right)^2lefrac{2k-1}{2k+1}
$$
Therefore,
$$
begin{align}
prod_{k=1}^nleft(frac{2k-1}{2k}right)^2
&leprod_{k=1}^nfrac{2k-1}{2k+1}\
&=frac1{2n+1}
end{align}
$$
Thus,
$$
prod_{k=1}^inftyfrac{2k-1}{2k}=0
$$
For $nge1$,
$$
begin{align}
frac{frac{sqrt{n^2-1}}n}{frac{sqrt{(n+1)^2-1}}{n+1}}
&=sqrt{frac{(n+1)^3(n-1)}{n^3(n+2)}}\
&=sqrt{frac{n^4+2n^3-2n-1}{n^4+2n^3}}\[18pt]
<1
end{align}
$$
Thus, $frac{sqrt{n^2-1}}n$ is increasing. Furthermore,
$$
begin{align}
lim_{ntoinfty}frac{sqrt{n^2-1}}n
&=lim_{ntoinfty}sqrt{1-frac1{n^2}}\[6pt]
&=1
end{align}
$$
add a comment |
Cross-multiplication yields, for $kge1$,
$$
left(frac{2k-1}{2k}right)^2lefrac{2k-1}{2k+1}
$$
Therefore,
$$
begin{align}
prod_{k=1}^nleft(frac{2k-1}{2k}right)^2
&leprod_{k=1}^nfrac{2k-1}{2k+1}\
&=frac1{2n+1}
end{align}
$$
Thus,
$$
prod_{k=1}^inftyfrac{2k-1}{2k}=0
$$
For $nge1$,
$$
begin{align}
frac{frac{sqrt{n^2-1}}n}{frac{sqrt{(n+1)^2-1}}{n+1}}
&=sqrt{frac{(n+1)^3(n-1)}{n^3(n+2)}}\
&=sqrt{frac{n^4+2n^3-2n-1}{n^4+2n^3}}\[18pt]
<1
end{align}
$$
Thus, $frac{sqrt{n^2-1}}n$ is increasing. Furthermore,
$$
begin{align}
lim_{ntoinfty}frac{sqrt{n^2-1}}n
&=lim_{ntoinfty}sqrt{1-frac1{n^2}}\[6pt]
&=1
end{align}
$$
add a comment |
Cross-multiplication yields, for $kge1$,
$$
left(frac{2k-1}{2k}right)^2lefrac{2k-1}{2k+1}
$$
Therefore,
$$
begin{align}
prod_{k=1}^nleft(frac{2k-1}{2k}right)^2
&leprod_{k=1}^nfrac{2k-1}{2k+1}\
&=frac1{2n+1}
end{align}
$$
Thus,
$$
prod_{k=1}^inftyfrac{2k-1}{2k}=0
$$
For $nge1$,
$$
begin{align}
frac{frac{sqrt{n^2-1}}n}{frac{sqrt{(n+1)^2-1}}{n+1}}
&=sqrt{frac{(n+1)^3(n-1)}{n^3(n+2)}}\
&=sqrt{frac{n^4+2n^3-2n-1}{n^4+2n^3}}\[18pt]
<1
end{align}
$$
Thus, $frac{sqrt{n^2-1}}n$ is increasing. Furthermore,
$$
begin{align}
lim_{ntoinfty}frac{sqrt{n^2-1}}n
&=lim_{ntoinfty}sqrt{1-frac1{n^2}}\[6pt]
&=1
end{align}
$$
Cross-multiplication yields, for $kge1$,
$$
left(frac{2k-1}{2k}right)^2lefrac{2k-1}{2k+1}
$$
Therefore,
$$
begin{align}
prod_{k=1}^nleft(frac{2k-1}{2k}right)^2
&leprod_{k=1}^nfrac{2k-1}{2k+1}\
&=frac1{2n+1}
end{align}
$$
Thus,
$$
prod_{k=1}^inftyfrac{2k-1}{2k}=0
$$
For $nge1$,
$$
begin{align}
frac{frac{sqrt{n^2-1}}n}{frac{sqrt{(n+1)^2-1}}{n+1}}
&=sqrt{frac{(n+1)^3(n-1)}{n^3(n+2)}}\
&=sqrt{frac{n^4+2n^3-2n-1}{n^4+2n^3}}\[18pt]
<1
end{align}
$$
Thus, $frac{sqrt{n^2-1}}n$ is increasing. Furthermore,
$$
begin{align}
lim_{ntoinfty}frac{sqrt{n^2-1}}n
&=lim_{ntoinfty}sqrt{1-frac1{n^2}}\[6pt]
&=1
end{align}
$$
edited Nov 25 at 4:35
answered Nov 24 at 22:50
robjohn♦
264k27302623
264k27302623
add a comment |
add a comment |
HINT
We have that
$$a_n=frac{1cdot 3cdot…cdot(2n-1)}{2cdot4cdot…cdot2n}=frac{(2n)!}{(2^nn!)^2}$$
then we can use Stirling's approximation.
Refer to the related
- What is the limit of $frac{prod Odd}{prod Even}?!$
Can you help with the other two?
– Sargis Iskandaryan
Nov 24 at 22:22
@SargisIskandaryan The second one is trivial indeed $frac{n^n}{n!}=frac{ncdot nldots n}{1cdot 2cdot n}$ or by ratio test for example.
– gimusi
Nov 24 at 22:28
@SargisIskandaryan I don't understand the first question.
– gimusi
Nov 24 at 22:29
add a comment |
HINT
We have that
$$a_n=frac{1cdot 3cdot…cdot(2n-1)}{2cdot4cdot…cdot2n}=frac{(2n)!}{(2^nn!)^2}$$
then we can use Stirling's approximation.
Refer to the related
- What is the limit of $frac{prod Odd}{prod Even}?!$
Can you help with the other two?
– Sargis Iskandaryan
Nov 24 at 22:22
@SargisIskandaryan The second one is trivial indeed $frac{n^n}{n!}=frac{ncdot nldots n}{1cdot 2cdot n}$ or by ratio test for example.
– gimusi
Nov 24 at 22:28
@SargisIskandaryan I don't understand the first question.
– gimusi
Nov 24 at 22:29
add a comment |
HINT
We have that
$$a_n=frac{1cdot 3cdot…cdot(2n-1)}{2cdot4cdot…cdot2n}=frac{(2n)!}{(2^nn!)^2}$$
then we can use Stirling's approximation.
Refer to the related
- What is the limit of $frac{prod Odd}{prod Even}?!$
HINT
We have that
$$a_n=frac{1cdot 3cdot…cdot(2n-1)}{2cdot4cdot…cdot2n}=frac{(2n)!}{(2^nn!)^2}$$
then we can use Stirling's approximation.
Refer to the related
- What is the limit of $frac{prod Odd}{prod Even}?!$
answered Nov 24 at 22:09
gimusi
1
1
Can you help with the other two?
– Sargis Iskandaryan
Nov 24 at 22:22
@SargisIskandaryan The second one is trivial indeed $frac{n^n}{n!}=frac{ncdot nldots n}{1cdot 2cdot n}$ or by ratio test for example.
– gimusi
Nov 24 at 22:28
@SargisIskandaryan I don't understand the first question.
– gimusi
Nov 24 at 22:29
add a comment |
Can you help with the other two?
– Sargis Iskandaryan
Nov 24 at 22:22
@SargisIskandaryan The second one is trivial indeed $frac{n^n}{n!}=frac{ncdot nldots n}{1cdot 2cdot n}$ or by ratio test for example.
– gimusi
Nov 24 at 22:28
@SargisIskandaryan I don't understand the first question.
– gimusi
Nov 24 at 22:29
Can you help with the other two?
– Sargis Iskandaryan
Nov 24 at 22:22
Can you help with the other two?
– Sargis Iskandaryan
Nov 24 at 22:22
@SargisIskandaryan The second one is trivial indeed $frac{n^n}{n!}=frac{ncdot nldots n}{1cdot 2cdot n}$ or by ratio test for example.
– gimusi
Nov 24 at 22:28
@SargisIskandaryan The second one is trivial indeed $frac{n^n}{n!}=frac{ncdot nldots n}{1cdot 2cdot n}$ or by ratio test for example.
– gimusi
Nov 24 at 22:28
@SargisIskandaryan I don't understand the first question.
– gimusi
Nov 24 at 22:29
@SargisIskandaryan I don't understand the first question.
– gimusi
Nov 24 at 22:29
add a comment |
$$frac{(2n-1)!!}{(2n)!!}=frac{1}{4^n}binom{2n}{n}=frac{2}{pi}int_{0}^{pi/2}left(costhetaright)^{2n},dtheta $$
is quite clearly decreasing and convergent to zero (by the dominated convergence theorem).
Laplace's method gives $frac{1}{4^n}binom{2n}{n}simfrac{1}{sqrt{pileft(n+frac{1}{4}right)}}$.
add a comment |
$$frac{(2n-1)!!}{(2n)!!}=frac{1}{4^n}binom{2n}{n}=frac{2}{pi}int_{0}^{pi/2}left(costhetaright)^{2n},dtheta $$
is quite clearly decreasing and convergent to zero (by the dominated convergence theorem).
Laplace's method gives $frac{1}{4^n}binom{2n}{n}simfrac{1}{sqrt{pileft(n+frac{1}{4}right)}}$.
add a comment |
$$frac{(2n-1)!!}{(2n)!!}=frac{1}{4^n}binom{2n}{n}=frac{2}{pi}int_{0}^{pi/2}left(costhetaright)^{2n},dtheta $$
is quite clearly decreasing and convergent to zero (by the dominated convergence theorem).
Laplace's method gives $frac{1}{4^n}binom{2n}{n}simfrac{1}{sqrt{pileft(n+frac{1}{4}right)}}$.
$$frac{(2n-1)!!}{(2n)!!}=frac{1}{4^n}binom{2n}{n}=frac{2}{pi}int_{0}^{pi/2}left(costhetaright)^{2n},dtheta $$
is quite clearly decreasing and convergent to zero (by the dominated convergence theorem).
Laplace's method gives $frac{1}{4^n}binom{2n}{n}simfrac{1}{sqrt{pileft(n+frac{1}{4}right)}}$.
answered Nov 24 at 22:24
Jack D'Aurizio
285k33278655
285k33278655
add a comment |
add a comment |
The sequence is decreasing and bounded below by 0. That's why it has a limit.
add a comment |
The sequence is decreasing and bounded below by 0. That's why it has a limit.
add a comment |
The sequence is decreasing and bounded below by 0. That's why it has a limit.
The sequence is decreasing and bounded below by 0. That's why it has a limit.
answered Nov 24 at 22:10
Martin Erhardt
1859
1859
add a comment |
add a comment |
1
If you have several questions to ask, it is better to ask separate questions, unless the questions are closely related.
– robjohn♦
Nov 25 at 1:34