Mixing
Proving that $lim_{k to infty} sqrt[k]{k!}$ diverges
$begingroup$
I am trying to prove that $lim_{k to infty} sqrt[k]{k!}$ diverges. I've rewritten this as $$
lim_{k to infty} expleft(frac{1}{k} sum_{n=1}^k log(n)right)
$$
which diverges if $lim_{k to infty} frac{1}{k} sum_{n=1}^k log(n)$ diverges
It is here that I am having trouble. I apply the ratio test and examine the limit $$
lim_{k to infty} left|
frac{log(k+1)/(k+1)}{log(k)/k}
right|=
lim_{k to infty} left|frac{log(k+1)}{log k}right|
left|frac{k}{k+1}right|.
$$
Letting $$
a_k = left|frac{log(k+1)}{log k}right|
~~text{ and }~~
b_k = left|frac{k}{k+1}right|,$$ it seems that $a_k$ is greater than 1.0 by an amount which is less than the amount by which $b_k$ is less than 1.
Therefore, it should be the case that $$
0 < lim_{ktoinfty} a_k b_k < 1
$$
which by the ratio tests would imply that the sum $frac{1}{k} sum_{n=1}^k log(n)$ converges.
Where am I going wrong?
real-analysis limits
asked Jan 31 at 23:18
tedted
596412
$endgroup$
|
show 1 more comment
$begingroup$
I am trying to prove that $lim_{k to infty} sqrt[k]{k!}$ diverges. I've rewritten this as $$
lim_{k to infty} expleft(frac{1}{k} sum_{n=1}^k log(n)right)
$$
which diverges if $lim_{k to infty} frac{1}{k} sum_{n=1}^k log(n)$ diverges
It is here that I am having trouble. I apply the ratio test and examine the limit $$
lim_{k to infty} left|
frac{log(k+1)/(k+1)}{log(k)/k}
right|=
lim_{k to infty} left|frac{log(k+1)}{log k}right|
left|frac{k}{k+1}right|.
$$
Letting $$
a_k = left|frac{log(k+1)}{log k}right|
~~text{ and }~~
b_k = left|frac{k}{k+1}right|,$$ it seems that $a_k$ is greater than 1.0 by an amount which is less than the amount by which $b_k$ is less than 1.
Therefore, it should be the case that $$
0 < lim_{ktoinfty} a_k b_k < 1
$$
which by the ratio tests would imply that the sum $frac{1}{k} sum_{n=1}^k log(n)$ converges.
Where am I going wrong?
real-analysis limits
asked Jan 31 at 23:18
tedted
596412
$endgroup$
$begingroup$
Do you know Stirling's formula?
$endgroup$
– Bernard
Jan 31 at 23:20
$begingroup$
To answer your question of "Where am I going wrong," the limit of $a_kb_k$ is equal to $1$.
$endgroup$
– Cheerful Parsnip
Jan 31 at 23:24
$begingroup$
@Bernard: thanks, Stirling's formula does the trick.
$endgroup$
– ted
Jan 31 at 23:28
$begingroup$
@Cheerful Parsnip: good point, I missed this. So in this case the ratio test would be inconclusive.
$endgroup$
– ted
Jan 31 at 23:28
$begingroup$
@ted: that's right.
$endgroup$
– Cheerful Parsnip
Jan 31 at 23:29
|
show 1 more comment
$begingroup$
I am trying to prove that $lim_{k to infty} sqrt[k]{k!}$ diverges. I've rewritten this as $$
lim_{k to infty} expleft(frac{1}{k} sum_{n=1}^k log(n)right)
$$
which diverges if $lim_{k to infty} frac{1}{k} sum_{n=1}^k log(n)$ diverges
It is here that I am having trouble. I apply the ratio test and examine the limit $$
lim_{k to infty} left|
frac{log(k+1)/(k+1)}{log(k)/k}
right|=
lim_{k to infty} left|frac{log(k+1)}{log k}right|
left|frac{k}{k+1}right|.
$$
Letting $$
a_k = left|frac{log(k+1)}{log k}right|
~~text{ and }~~
b_k = left|frac{k}{k+1}right|,$$ it seems that $a_k$ is greater than 1.0 by an amount which is less than the amount by which $b_k$ is less than 1.
Therefore, it should be the case that $$
0 < lim_{ktoinfty} a_k b_k < 1
$$
which by the ratio tests would imply that the sum $frac{1}{k} sum_{n=1}^k log(n)$ converges.
Where am I going wrong?
real-analysis limits
asked Jan 31 at 23:18
tedted
596412
$endgroup$
I am trying to prove that $lim_{k to infty} sqrt[k]{k!}$ diverges. I've rewritten this as $$
lim_{k to infty} expleft(frac{1}{k} sum_{n=1}^k log(n)right)
$$
which diverges if $lim_{k to infty} frac{1}{k} sum_{n=1}^k log(n)$ diverges
It is here that I am having trouble. I apply the ratio test and examine the limit $$
lim_{k to infty} left|
frac{log(k+1)/(k+1)}{log(k)/k}
right|=
lim_{k to infty} left|frac{log(k+1)}{log k}right|
left|frac{k}{k+1}right|.
$$
Letting $$
a_k = left|frac{log(k+1)}{log k}right|
~~text{ and }~~
b_k = left|frac{k}{k+1}right|,$$ it seems that $a_k$ is greater than 1.0 by an amount which is less than the amount by which $b_k$ is less than 1.
Therefore, it should be the case that $$
0 < lim_{ktoinfty} a_k b_k < 1
$$
which by the ratio tests would imply that the sum $frac{1}{k} sum_{n=1}^k log(n)$ converges.
Where am I going wrong?
real-analysis limits
real-analysis limits
asked Jan 31 at 23:18
tedted
596412
asked Jan 31 at 23:18
tedted
596412
asked Jan 31 at 23:18
tedted
596412
asked Jan 31 at 23:18
tedted
596412
asked Jan 31 at 23:18
tedted
596412
596412
$begingroup$
Do you know Stirling's formula?
$endgroup$
– Bernard
Jan 31 at 23:20
$begingroup$
To answer your question of "Where am I going wrong," the limit of $a_kb_k$ is equal to $1$.
$endgroup$
– Cheerful Parsnip
Jan 31 at 23:24
$begingroup$
@Bernard: thanks, Stirling's formula does the trick.
$endgroup$
– ted
Jan 31 at 23:28
$begingroup$
@Cheerful Parsnip: good point, I missed this. So in this case the ratio test would be inconclusive.
$endgroup$
– ted
Jan 31 at 23:28
$begingroup$
@ted: that's right.
$endgroup$
– Cheerful Parsnip
Jan 31 at 23:29
|
show 1 more comment
$begingroup$
Do you know Stirling's formula?
$endgroup$
– Bernard
Jan 31 at 23:20
$begingroup$
To answer your question of "Where am I going wrong," the limit of $a_kb_k$ is equal to $1$.
$endgroup$
– Cheerful Parsnip
Jan 31 at 23:24
$begingroup$
@Bernard: thanks, Stirling's formula does the trick.
$endgroup$
– ted
Jan 31 at 23:28
$begingroup$
@Cheerful Parsnip: good point, I missed this. So in this case the ratio test would be inconclusive.
$endgroup$
– ted
Jan 31 at 23:28
$begingroup$
@ted: that's right.
$endgroup$
– Cheerful Parsnip
Jan 31 at 23:29
$begingroup$
Do you know Stirling's formula?
$endgroup$
– Bernard
Jan 31 at 23:20
$begingroup$
Do you know Stirling's formula?
$endgroup$
– Bernard
Jan 31 at 23:20
$begingroup$
To answer your question of "Where am I going wrong," the limit of $a_kb_k$ is equal to $1$.
$endgroup$
– Cheerful Parsnip
Jan 31 at 23:24
$begingroup$
To answer your question of "Where am I going wrong," the limit of $a_kb_k$ is equal to $1$.
$endgroup$
– Cheerful Parsnip
Jan 31 at 23:24
$begingroup$
@Bernard: thanks, Stirling's formula does the trick.
$endgroup$
– ted
Jan 31 at 23:28
$begingroup$
@Bernard: thanks, Stirling's formula does the trick.
$endgroup$
– ted
Jan 31 at 23:28
$begingroup$
@Cheerful Parsnip: good point, I missed this. So in this case the ratio test would be inconclusive.
$endgroup$
– ted
Jan 31 at 23:28
$begingroup$
@Cheerful Parsnip: good point, I missed this. So in this case the ratio test would be inconclusive.
$endgroup$
– ted
Jan 31 at 23:28
$begingroup$
@ted: that's right.
$endgroup$
– Cheerful Parsnip
Jan 31 at 23:29
$begingroup$
@ted: that's right.
$endgroup$
– Cheerful Parsnip
Jan 31 at 23:29
|
show 1 more comment
3 Answers
3
active
oldest
votes
$begingroup$
"it seems that $a_k$ is greater than 1.0 by an amount which is less than the amount by which $b_k$ is less than 1"
This is where you are going wrong. Regardless of whether this somewhat hazy statement is true (for some notion of "amount", etc.), the ratio test does not care about these considerations. Both quantities go to 1: thus the ratio goes to 1. A limit of 1 for the ratio leads to an inconclusive ratio test.
Therefore, you need to use a different, or more fine-grained approach: you cannot conclude by the ratio test.
As discussed in the comments, you may want to use refinements of the ratio test to try and conclude with this type of argument. Raabe's test (first refinement: look at $lim_{ktoinfty} k(frac{a_k}{a_{k+1}}-1)$) will still be inconclusive, but Bertrand's test (stronger refinement: look at $lim_{ktoinfty} left(klog k (frac{a_k}{a_{k+1}}-1) - log kright)$) will let you establish the result.
answered Jan 31 at 23:24
Clement C.Clement C.
51.1k34093
$endgroup$
$begingroup$
For what it's worth, by the way, your hazy statement is true: the ratio behaves like $$1-1/k + o(1/k)$$ as $ktoinfty$. But this still leads to an inconclusive ratio test (and Raabe's test as well, for that matter). (But Bertrand's test, a finer extension, would allow you to conclude)
$endgroup$
– Clement C.
Jan 31 at 23:29
add a comment |
$begingroup$
Hint: $k!ge (k/2)^{k/2}$ for large $k.$
answered Jan 31 at 23:21
zhw.zhw.
74.9k43275
$endgroup$
$begingroup$
This is true, but it doesn't actually answer the question {"Where am I going wrong?")
$endgroup$
– Clement C.
Jan 31 at 23:38
add a comment |
$begingroup$
Let $L=lim limits_{k to infty} sqrt[k] k!$.We have $ln L=lim limits_{k to infty} ln sqrt[k] k!=lim limits_{k to infty} frac{ln k!}{k} $.
By Stolz-Cesaro's Lemma,$lim limits_{k to infty} frac{ln k!}{k}=lim limits_{k to infty}ln(k+1) =infty$,so $L=infty$.
answered Jan 31 at 23:24
AlexdanutAlexdanut
1738
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
"it seems that $a_k$ is greater than 1.0 by an amount which is less than the amount by which $b_k$ is less than 1"
This is where you are going wrong. Regardless of whether this somewhat hazy statement is true (for some notion of "amount", etc.), the ratio test does not care about these considerations. Both quantities go to 1: thus the ratio goes to 1. A limit of 1 for the ratio leads to an inconclusive ratio test.
Therefore, you need to use a different, or more fine-grained approach: you cannot conclude by the ratio test.
As discussed in the comments, you may want to use refinements of the ratio test to try and conclude with this type of argument. Raabe's test (first refinement: look at $lim_{ktoinfty} k(frac{a_k}{a_{k+1}}-1)$) will still be inconclusive, but Bertrand's test (stronger refinement: look at $lim_{ktoinfty} left(klog k (frac{a_k}{a_{k+1}}-1) - log kright)$) will let you establish the result.
answered Jan 31 at 23:24
Clement C.Clement C.
51.1k34093
$endgroup$
$begingroup$
For what it's worth, by the way, your hazy statement is true: the ratio behaves like $$1-1/k + o(1/k)$$ as $ktoinfty$. But this still leads to an inconclusive ratio test (and Raabe's test as well, for that matter). (But Bertrand's test, a finer extension, would allow you to conclude)
$endgroup$
– Clement C.
Jan 31 at 23:29
add a comment |
$begingroup$
"it seems that $a_k$ is greater than 1.0 by an amount which is less than the amount by which $b_k$ is less than 1"
This is where you are going wrong. Regardless of whether this somewhat hazy statement is true (for some notion of "amount", etc.), the ratio test does not care about these considerations. Both quantities go to 1: thus the ratio goes to 1. A limit of 1 for the ratio leads to an inconclusive ratio test.
Therefore, you need to use a different, or more fine-grained approach: you cannot conclude by the ratio test.
As discussed in the comments, you may want to use refinements of the ratio test to try and conclude with this type of argument. Raabe's test (first refinement: look at $lim_{ktoinfty} k(frac{a_k}{a_{k+1}}-1)$) will still be inconclusive, but Bertrand's test (stronger refinement: look at $lim_{ktoinfty} left(klog k (frac{a_k}{a_{k+1}}-1) - log kright)$) will let you establish the result.
answered Jan 31 at 23:24
Clement C.Clement C.
51.1k34093
$endgroup$
$begingroup$
For what it's worth, by the way, your hazy statement is true: the ratio behaves like $$1-1/k + o(1/k)$$ as $ktoinfty$. But this still leads to an inconclusive ratio test (and Raabe's test as well, for that matter). (But Bertrand's test, a finer extension, would allow you to conclude)
$endgroup$
– Clement C.
Jan 31 at 23:29
add a comment |
$begingroup$
"it seems that $a_k$ is greater than 1.0 by an amount which is less than the amount by which $b_k$ is less than 1"
This is where you are going wrong. Regardless of whether this somewhat hazy statement is true (for some notion of "amount", etc.), the ratio test does not care about these considerations. Both quantities go to 1: thus the ratio goes to 1. A limit of 1 for the ratio leads to an inconclusive ratio test.
Therefore, you need to use a different, or more fine-grained approach: you cannot conclude by the ratio test.
As discussed in the comments, you may want to use refinements of the ratio test to try and conclude with this type of argument. Raabe's test (first refinement: look at $lim_{ktoinfty} k(frac{a_k}{a_{k+1}}-1)$) will still be inconclusive, but Bertrand's test (stronger refinement: look at $lim_{ktoinfty} left(klog k (frac{a_k}{a_{k+1}}-1) - log kright)$) will let you establish the result.
answered Jan 31 at 23:24
Clement C.Clement C.
51.1k34093
$endgroup$
"it seems that $a_k$ is greater than 1.0 by an amount which is less than the amount by which $b_k$ is less than 1"
This is where you are going wrong. Regardless of whether this somewhat hazy statement is true (for some notion of "amount", etc.), the ratio test does not care about these considerations. Both quantities go to 1: thus the ratio goes to 1. A limit of 1 for the ratio leads to an inconclusive ratio test.
Therefore, you need to use a different, or more fine-grained approach: you cannot conclude by the ratio test.
As discussed in the comments, you may want to use refinements of the ratio test to try and conclude with this type of argument. Raabe's test (first refinement: look at $lim_{ktoinfty} k(frac{a_k}{a_{k+1}}-1)$) will still be inconclusive, but Bertrand's test (stronger refinement: look at $lim_{ktoinfty} left(klog k (frac{a_k}{a_{k+1}}-1) - log kright)$) will let you establish the result.
answered Jan 31 at 23:24
Clement C.Clement C.
51.1k34093
edited Jan 31 at 23:40
answered Jan 31 at 23:24
Clement C.Clement C.
51.1k34093
answered Jan 31 at 23:24
Clement C.Clement C.
51.1k34093
answered Jan 31 at 23:24
Clement C.Clement C.
51.1k34093
51.1k34093
$begingroup$
For what it's worth, by the way, your hazy statement is true: the ratio behaves like $$1-1/k + o(1/k)$$ as $ktoinfty$. But this still leads to an inconclusive ratio test (and Raabe's test as well, for that matter). (But Bertrand's test, a finer extension, would allow you to conclude)
$endgroup$
– Clement C.
Jan 31 at 23:29
add a comment |
$begingroup$
For what it's worth, by the way, your hazy statement is true: the ratio behaves like $$1-1/k + o(1/k)$$ as $ktoinfty$. But this still leads to an inconclusive ratio test (and Raabe's test as well, for that matter). (But Bertrand's test, a finer extension, would allow you to conclude)
$endgroup$
– Clement C.
Jan 31 at 23:29
$begingroup$
For what it's worth, by the way, your hazy statement is true: the ratio behaves like $$1-1/k + o(1/k)$$ as $ktoinfty$. But this still leads to an inconclusive ratio test (and Raabe's test as well, for that matter). (But Bertrand's test, a finer extension, would allow you to conclude)
$endgroup$
– Clement C.
Jan 31 at 23:29
$begingroup$
For what it's worth, by the way, your hazy statement is true: the ratio behaves like $$1-1/k + o(1/k)$$ as $ktoinfty$. But this still leads to an inconclusive ratio test (and Raabe's test as well, for that matter). (But Bertrand's test, a finer extension, would allow you to conclude)
$endgroup$
– Clement C.
Jan 31 at 23:29
add a comment |
$begingroup$
Hint: $k!ge (k/2)^{k/2}$ for large $k.$
answered Jan 31 at 23:21
zhw.zhw.
74.9k43275
$endgroup$
$begingroup$
This is true, but it doesn't actually answer the question {"Where am I going wrong?")
$endgroup$
– Clement C.
Jan 31 at 23:38
add a comment |
$begingroup$
Hint: $k!ge (k/2)^{k/2}$ for large $k.$
answered Jan 31 at 23:21
zhw.zhw.
74.9k43275
$endgroup$
$begingroup$
This is true, but it doesn't actually answer the question {"Where am I going wrong?")
$endgroup$
– Clement C.
Jan 31 at 23:38
add a comment |
$begingroup$
Hint: $k!ge (k/2)^{k/2}$ for large $k.$
answered Jan 31 at 23:21
zhw.zhw.
74.9k43275
$endgroup$
Hint: $k!ge (k/2)^{k/2}$ for large $k.$
answered Jan 31 at 23:21
zhw.zhw.
74.9k43275
answered Jan 31 at 23:21
zhw.zhw.
74.9k43275
answered Jan 31 at 23:21
zhw.zhw.
74.9k43275
answered Jan 31 at 23:21
zhw.zhw.
74.9k43275
74.9k43275
$begingroup$
This is true, but it doesn't actually answer the question {"Where am I going wrong?")
$endgroup$
– Clement C.
Jan 31 at 23:38
add a comment |
$begingroup$
This is true, but it doesn't actually answer the question {"Where am I going wrong?")
$endgroup$
– Clement C.
Jan 31 at 23:38
$begingroup$
This is true, but it doesn't actually answer the question {"Where am I going wrong?")
$endgroup$
– Clement C.
Jan 31 at 23:38
$begingroup$
This is true, but it doesn't actually answer the question {"Where am I going wrong?")
$endgroup$
– Clement C.
Jan 31 at 23:38
add a comment |
$begingroup$
Let $L=lim limits_{k to infty} sqrt[k] k!$.We have $ln L=lim limits_{k to infty} ln sqrt[k] k!=lim limits_{k to infty} frac{ln k!}{k} $.
By Stolz-Cesaro's Lemma,$lim limits_{k to infty} frac{ln k!}{k}=lim limits_{k to infty}ln(k+1) =infty$,so $L=infty$.
answered Jan 31 at 23:24
AlexdanutAlexdanut
1738
$endgroup$
add a comment |
$begingroup$
Let $L=lim limits_{k to infty} sqrt[k] k!$.We have $ln L=lim limits_{k to infty} ln sqrt[k] k!=lim limits_{k to infty} frac{ln k!}{k} $.
By Stolz-Cesaro's Lemma,$lim limits_{k to infty} frac{ln k!}{k}=lim limits_{k to infty}ln(k+1) =infty$,so $L=infty$.
answered Jan 31 at 23:24
AlexdanutAlexdanut
1738
$endgroup$
add a comment |
$begingroup$
Let $L=lim limits_{k to infty} sqrt[k] k!$.We have $ln L=lim limits_{k to infty} ln sqrt[k] k!=lim limits_{k to infty} frac{ln k!}{k} $.
By Stolz-Cesaro's Lemma,$lim limits_{k to infty} frac{ln k!}{k}=lim limits_{k to infty}ln(k+1) =infty$,so $L=infty$.
answered Jan 31 at 23:24
AlexdanutAlexdanut
1738
$endgroup$
Let $L=lim limits_{k to infty} sqrt[k] k!$.We have $ln L=lim limits_{k to infty} ln sqrt[k] k!=lim limits_{k to infty} frac{ln k!}{k} $.
By Stolz-Cesaro's Lemma,$lim limits_{k to infty} frac{ln k!}{k}=lim limits_{k to infty}ln(k+1) =infty$,so $L=infty$.
answered Jan 31 at 23:24
AlexdanutAlexdanut
1738
answered Jan 31 at 23:24
AlexdanutAlexdanut
1738
answered Jan 31 at 23:24
AlexdanutAlexdanut
1738
answered Jan 31 at 23:24
AlexdanutAlexdanut
1738
1738
add a comment |
add a comment |
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$begingroup$
Do you know Stirling's formula?
$endgroup$
– Bernard
Jan 31 at 23:20
$begingroup$
To answer your question of "Where am I going wrong," the limit of $a_kb_k$ is equal to $1$.
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– Cheerful Parsnip
Jan 31 at 23:24
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@Bernard: thanks, Stirling's formula does the trick.
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– ted
Jan 31 at 23:28
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@Cheerful Parsnip: good point, I missed this. So in this case the ratio test would be inconclusive.
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– ted
Jan 31 at 23:28
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@ted: that's right.
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– Cheerful Parsnip
Jan 31 at 23:29