Theorem 3.37 in Baby Rudin: Any Counter-Examples In The Other Case?
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Here is Theorem 3.37 in the book Principles Of Mathematical Analysis by Walter Rudin, 3rd edition:
For any sequence $left{ c_n right}$ of positive numbers,
$$ liminf_{ntoinfty} frac{c_{n+1}}{c_n} leq liminf_{ntoinfty} sqrt[n]{c_n} leq limsup_{ntoinfty} sqrt[n]{c_n} leq limsup_{ntoinfty} frac{c_{n+1}}{c_n}. $$
I think I fully understand the proof by Rudin.
From Theorem 3.37 in Baby Rudin, we can also conclude that following:
For any sequence $left{ c_n right}$ of positive numbers, if the sequence $left{ frac{c_{n+1}}{c_n} right}$ converges in $mathbb{R}$, then so does the sequence $left{ sqrt[n]{c_n} right}$, and then the two limits are equal.
Am I right?
However, I'm unable to figure out the proof of or come up with any counter-examples to the following:
Suppose that $left{ c_n right}$ is a sequence of positive real numbers such that the sequence $left{ sqrt[n]{c_n} right}$ converges in $mathbb{R}$. Then so does the sequence $left{ frac{c_{n+1}}{c_n} right}$.
What if the sequence $left{ sqrt[n]{c_n} right}$ converges in $mathbb{R} cup { pm infty }$? Does the sequence $left{ frac{c_{n+1}}{c_n} right}$ then also converge in $mathbb{R} cup { pm infty }$?
real-analysis sequences-and-series limits analysis limsup-and-liminf
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add a comment |
$begingroup$
Here is Theorem 3.37 in the book Principles Of Mathematical Analysis by Walter Rudin, 3rd edition:
For any sequence $left{ c_n right}$ of positive numbers,
$$ liminf_{ntoinfty} frac{c_{n+1}}{c_n} leq liminf_{ntoinfty} sqrt[n]{c_n} leq limsup_{ntoinfty} sqrt[n]{c_n} leq limsup_{ntoinfty} frac{c_{n+1}}{c_n}. $$
I think I fully understand the proof by Rudin.
From Theorem 3.37 in Baby Rudin, we can also conclude that following:
For any sequence $left{ c_n right}$ of positive numbers, if the sequence $left{ frac{c_{n+1}}{c_n} right}$ converges in $mathbb{R}$, then so does the sequence $left{ sqrt[n]{c_n} right}$, and then the two limits are equal.
Am I right?
However, I'm unable to figure out the proof of or come up with any counter-examples to the following:
Suppose that $left{ c_n right}$ is a sequence of positive real numbers such that the sequence $left{ sqrt[n]{c_n} right}$ converges in $mathbb{R}$. Then so does the sequence $left{ frac{c_{n+1}}{c_n} right}$.
What if the sequence $left{ sqrt[n]{c_n} right}$ converges in $mathbb{R} cup { pm infty }$? Does the sequence $left{ frac{c_{n+1}}{c_n} right}$ then also converge in $mathbb{R} cup { pm infty }$?
real-analysis sequences-and-series limits analysis limsup-and-liminf
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1
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math.stackexchange.com/a/1708611/43949
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– angryavian
Dec 3 '18 at 5:34
2
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Think as $sqrt[n]{c_n}$ as (almost) the geometric mean of $c_{i+1}/c_i$ for $i=0..n$. Applying the $log$ function, we can think of this additively instead. Then the question is to find a sequence of real numbers not converging such that the Cesàro mean converges…
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– Idéophage
Dec 3 '18 at 5:42
add a comment |
$begingroup$
Here is Theorem 3.37 in the book Principles Of Mathematical Analysis by Walter Rudin, 3rd edition:
For any sequence $left{ c_n right}$ of positive numbers,
$$ liminf_{ntoinfty} frac{c_{n+1}}{c_n} leq liminf_{ntoinfty} sqrt[n]{c_n} leq limsup_{ntoinfty} sqrt[n]{c_n} leq limsup_{ntoinfty} frac{c_{n+1}}{c_n}. $$
I think I fully understand the proof by Rudin.
From Theorem 3.37 in Baby Rudin, we can also conclude that following:
For any sequence $left{ c_n right}$ of positive numbers, if the sequence $left{ frac{c_{n+1}}{c_n} right}$ converges in $mathbb{R}$, then so does the sequence $left{ sqrt[n]{c_n} right}$, and then the two limits are equal.
Am I right?
However, I'm unable to figure out the proof of or come up with any counter-examples to the following:
Suppose that $left{ c_n right}$ is a sequence of positive real numbers such that the sequence $left{ sqrt[n]{c_n} right}$ converges in $mathbb{R}$. Then so does the sequence $left{ frac{c_{n+1}}{c_n} right}$.
What if the sequence $left{ sqrt[n]{c_n} right}$ converges in $mathbb{R} cup { pm infty }$? Does the sequence $left{ frac{c_{n+1}}{c_n} right}$ then also converge in $mathbb{R} cup { pm infty }$?
real-analysis sequences-and-series limits analysis limsup-and-liminf
$endgroup$
Here is Theorem 3.37 in the book Principles Of Mathematical Analysis by Walter Rudin, 3rd edition:
For any sequence $left{ c_n right}$ of positive numbers,
$$ liminf_{ntoinfty} frac{c_{n+1}}{c_n} leq liminf_{ntoinfty} sqrt[n]{c_n} leq limsup_{ntoinfty} sqrt[n]{c_n} leq limsup_{ntoinfty} frac{c_{n+1}}{c_n}. $$
I think I fully understand the proof by Rudin.
From Theorem 3.37 in Baby Rudin, we can also conclude that following:
For any sequence $left{ c_n right}$ of positive numbers, if the sequence $left{ frac{c_{n+1}}{c_n} right}$ converges in $mathbb{R}$, then so does the sequence $left{ sqrt[n]{c_n} right}$, and then the two limits are equal.
Am I right?
However, I'm unable to figure out the proof of or come up with any counter-examples to the following:
Suppose that $left{ c_n right}$ is a sequence of positive real numbers such that the sequence $left{ sqrt[n]{c_n} right}$ converges in $mathbb{R}$. Then so does the sequence $left{ frac{c_{n+1}}{c_n} right}$.
What if the sequence $left{ sqrt[n]{c_n} right}$ converges in $mathbb{R} cup { pm infty }$? Does the sequence $left{ frac{c_{n+1}}{c_n} right}$ then also converge in $mathbb{R} cup { pm infty }$?
real-analysis sequences-and-series limits analysis limsup-and-liminf
real-analysis sequences-and-series limits analysis limsup-and-liminf
asked Dec 3 '18 at 5:27
Saaqib MahmoodSaaqib Mahmood
7,71242477
7,71242477
1
$begingroup$
math.stackexchange.com/a/1708611/43949
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– angryavian
Dec 3 '18 at 5:34
2
$begingroup$
Think as $sqrt[n]{c_n}$ as (almost) the geometric mean of $c_{i+1}/c_i$ for $i=0..n$. Applying the $log$ function, we can think of this additively instead. Then the question is to find a sequence of real numbers not converging such that the Cesàro mean converges…
$endgroup$
– Idéophage
Dec 3 '18 at 5:42
add a comment |
1
$begingroup$
math.stackexchange.com/a/1708611/43949
$endgroup$
– angryavian
Dec 3 '18 at 5:34
2
$begingroup$
Think as $sqrt[n]{c_n}$ as (almost) the geometric mean of $c_{i+1}/c_i$ for $i=0..n$. Applying the $log$ function, we can think of this additively instead. Then the question is to find a sequence of real numbers not converging such that the Cesàro mean converges…
$endgroup$
– Idéophage
Dec 3 '18 at 5:42
1
1
$begingroup$
math.stackexchange.com/a/1708611/43949
$endgroup$
– angryavian
Dec 3 '18 at 5:34
$begingroup$
math.stackexchange.com/a/1708611/43949
$endgroup$
– angryavian
Dec 3 '18 at 5:34
2
2
$begingroup$
Think as $sqrt[n]{c_n}$ as (almost) the geometric mean of $c_{i+1}/c_i$ for $i=0..n$. Applying the $log$ function, we can think of this additively instead. Then the question is to find a sequence of real numbers not converging such that the Cesàro mean converges…
$endgroup$
– Idéophage
Dec 3 '18 at 5:42
$begingroup$
Think as $sqrt[n]{c_n}$ as (almost) the geometric mean of $c_{i+1}/c_i$ for $i=0..n$. Applying the $log$ function, we can think of this additively instead. Then the question is to find a sequence of real numbers not converging such that the Cesàro mean converges…
$endgroup$
– Idéophage
Dec 3 '18 at 5:42
add a comment |
1 Answer
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Concerning the first question: yes, you are right.
On the other hand, consider the sequence$$1,1,frac12,frac12,frac14,frac14,ldots$$In the case, $lim_{ntoinfty}sqrt[n]{c_n}=dfrac1{sqrt2}$, but $lim_{ntoinfty}dfrac{c_{n+1}}{c_n}$ doesn't exist.
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Concerning the first question: yes, you are right.
On the other hand, consider the sequence$$1,1,frac12,frac12,frac14,frac14,ldots$$In the case, $lim_{ntoinfty}sqrt[n]{c_n}=dfrac1{sqrt2}$, but $lim_{ntoinfty}dfrac{c_{n+1}}{c_n}$ doesn't exist.
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add a comment |
$begingroup$
Concerning the first question: yes, you are right.
On the other hand, consider the sequence$$1,1,frac12,frac12,frac14,frac14,ldots$$In the case, $lim_{ntoinfty}sqrt[n]{c_n}=dfrac1{sqrt2}$, but $lim_{ntoinfty}dfrac{c_{n+1}}{c_n}$ doesn't exist.
$endgroup$
add a comment |
$begingroup$
Concerning the first question: yes, you are right.
On the other hand, consider the sequence$$1,1,frac12,frac12,frac14,frac14,ldots$$In the case, $lim_{ntoinfty}sqrt[n]{c_n}=dfrac1{sqrt2}$, but $lim_{ntoinfty}dfrac{c_{n+1}}{c_n}$ doesn't exist.
$endgroup$
Concerning the first question: yes, you are right.
On the other hand, consider the sequence$$1,1,frac12,frac12,frac14,frac14,ldots$$In the case, $lim_{ntoinfty}sqrt[n]{c_n}=dfrac1{sqrt2}$, but $lim_{ntoinfty}dfrac{c_{n+1}}{c_n}$ doesn't exist.
answered Dec 3 '18 at 6:57
José Carlos SantosJosé Carlos Santos
157k22126227
157k22126227
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$begingroup$
math.stackexchange.com/a/1708611/43949
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– angryavian
Dec 3 '18 at 5:34
2
$begingroup$
Think as $sqrt[n]{c_n}$ as (almost) the geometric mean of $c_{i+1}/c_i$ for $i=0..n$. Applying the $log$ function, we can think of this additively instead. Then the question is to find a sequence of real numbers not converging such that the Cesàro mean converges…
$endgroup$
– Idéophage
Dec 3 '18 at 5:42