Proving the limit of x to infinity of x!/x^x is 0 [duplicate]
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This question already has an answer here:
What's the limit of the sequence $limlimits_{n toinfty} frac{n!}{n^n}$?
2 answers
I have found that the limit of x to infinity of x!/x^x is zero by doing this:
x^x = xxxx...(x amount of times)
x! = x(x-1)(x-2)(x-3)...(x amount of times)
So obviously, x^x grows much faster, because x is greater than x-1 and x-2 and x-3 and all those other terms. Therefore the limit must be 0.
But how to actually prove this with equations. And why does Wolfram Alpha disagree with me when I do the infinite summation of x!/x^x and it says by the limit test that the summation diverges:
https://www.wolframalpha.com/input/?i=sum+x!%2Fx%5Ex,+j%3D0..infinity
calculus limits
asked Jan 2 at 19:32
Math BobMath Bob
8111
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marked as duplicate by Martin R, Arthur, Lee Mosher, JMoravitz, Calvin Khor Jan 2 at 19:52
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:
What's the limit of the sequence $limlimits_{n toinfty} frac{n!}{n^n}$?
2 answers
I have found that the limit of x to infinity of x!/x^x is zero by doing this:
x^x = xxxx...(x amount of times)
x! = x(x-1)(x-2)(x-3)...(x amount of times)
So obviously, x^x grows much faster, because x is greater than x-1 and x-2 and x-3 and all those other terms. Therefore the limit must be 0.
But how to actually prove this with equations. And why does Wolfram Alpha disagree with me when I do the infinite summation of x!/x^x and it says by the limit test that the summation diverges:
https://www.wolframalpha.com/input/?i=sum+x!%2Fx%5Ex,+j%3D0..infinity
calculus limits
asked Jan 2 at 19:32
Math BobMath Bob
8111
$endgroup$
marked as duplicate by Martin R, Arthur, Lee Mosher, JMoravitz, Calvin Khor Jan 2 at 19:52
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
$begingroup$
The limit of a sequence and the sum of a series are different things ...
$endgroup$
– Martin R
Jan 2 at 19:34
$begingroup$
Are you calculating the limit of that sequence or the series of the sequence??
$endgroup$
– Math_QED
Jan 2 at 19:34
$begingroup$
Regarding the WA link, note that you write x!/x^x, but sum over j...
$endgroup$
– Wojowu
Jan 2 at 19:35
$begingroup$
The reason that wolfram alpha says your sum diverges is as Wojowu says you used the variable $x$ instead of $j$ in that sum. On top of this you need to start the index at $j=1$ on wolfram because of its convention with $0^0$. you can see this here: wolframalpha.com/input/?i=sum+j!%2Fj%5Ej,+j%3D1..infinity Contrary to the answer youve accepted the sum actually does converge, and you can show this pretty easily with a comparison with $sum frac{2}{n^2}$
$endgroup$
– B.Martin
Jan 3 at 0:19
add a comment |
$begingroup$
This question already has an answer here:
What's the limit of the sequence $limlimits_{n toinfty} frac{n!}{n^n}$?
2 answers
I have found that the limit of x to infinity of x!/x^x is zero by doing this:
x^x = xxxx...(x amount of times)
x! = x(x-1)(x-2)(x-3)...(x amount of times)
So obviously, x^x grows much faster, because x is greater than x-1 and x-2 and x-3 and all those other terms. Therefore the limit must be 0.
But how to actually prove this with equations. And why does Wolfram Alpha disagree with me when I do the infinite summation of x!/x^x and it says by the limit test that the summation diverges:
https://www.wolframalpha.com/input/?i=sum+x!%2Fx%5Ex,+j%3D0..infinity
calculus limits
asked Jan 2 at 19:32
Math BobMath Bob
8111
$endgroup$
This question already has an answer here:
What's the limit of the sequence $limlimits_{n toinfty} frac{n!}{n^n}$?
2 answers
I have found that the limit of x to infinity of x!/x^x is zero by doing this:
x^x = xxxx...(x amount of times)
x! = x(x-1)(x-2)(x-3)...(x amount of times)
So obviously, x^x grows much faster, because x is greater than x-1 and x-2 and x-3 and all those other terms. Therefore the limit must be 0.
But how to actually prove this with equations. And why does Wolfram Alpha disagree with me when I do the infinite summation of x!/x^x and it says by the limit test that the summation diverges:
https://www.wolframalpha.com/input/?i=sum+x!%2Fx%5Ex,+j%3D0..infinity
This question already has an answer here:
What's the limit of the sequence $limlimits_{n toinfty} frac{n!}{n^n}$?
2 answers
calculus limits
calculus limits
asked Jan 2 at 19:32
Math BobMath Bob
8111
asked Jan 2 at 19:32
Math BobMath Bob
8111
asked Jan 2 at 19:32
Math BobMath Bob
8111
asked Jan 2 at 19:32
Math BobMath Bob
8111
asked Jan 2 at 19:32
Math BobMath Bob
8111
8111
marked as duplicate by Martin R, Arthur, Lee Mosher, JMoravitz, Calvin Khor Jan 2 at 19:52
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 Martin R, Arthur, Lee Mosher, JMoravitz, Calvin Khor Jan 2 at 19:52
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
$begingroup$
The limit of a sequence and the sum of a series are different things ...
$endgroup$
– Martin R
Jan 2 at 19:34
$begingroup$
Are you calculating the limit of that sequence or the series of the sequence??
$endgroup$
– Math_QED
Jan 2 at 19:34
$begingroup$
Regarding the WA link, note that you write x!/x^x, but sum over j...
$endgroup$
– Wojowu
Jan 2 at 19:35
$begingroup$
The reason that wolfram alpha says your sum diverges is as Wojowu says you used the variable $x$ instead of $j$ in that sum. On top of this you need to start the index at $j=1$ on wolfram because of its convention with $0^0$. you can see this here: wolframalpha.com/input/?i=sum+j!%2Fj%5Ej,+j%3D1..infinity Contrary to the answer youve accepted the sum actually does converge, and you can show this pretty easily with a comparison with $sum frac{2}{n^2}$
$endgroup$
– B.Martin
Jan 3 at 0:19
add a comment |
1
$begingroup$
The limit of a sequence and the sum of a series are different things ...
$endgroup$
– Martin R
Jan 2 at 19:34
$begingroup$
Are you calculating the limit of that sequence or the series of the sequence??
$endgroup$
– Math_QED
Jan 2 at 19:34
$begingroup$
Regarding the WA link, note that you write x!/x^x, but sum over j...
$endgroup$
– Wojowu
Jan 2 at 19:35
$begingroup$
The reason that wolfram alpha says your sum diverges is as Wojowu says you used the variable $x$ instead of $j$ in that sum. On top of this you need to start the index at $j=1$ on wolfram because of its convention with $0^0$. you can see this here: wolframalpha.com/input/?i=sum+j!%2Fj%5Ej,+j%3D1..infinity Contrary to the answer youve accepted the sum actually does converge, and you can show this pretty easily with a comparison with $sum frac{2}{n^2}$
$endgroup$
– B.Martin
Jan 3 at 0:19
1
1
$begingroup$
The limit of a sequence and the sum of a series are different things ...
$endgroup$
– Martin R
Jan 2 at 19:34
$begingroup$
The limit of a sequence and the sum of a series are different things ...
$endgroup$
– Martin R
Jan 2 at 19:34
$begingroup$
Are you calculating the limit of that sequence or the series of the sequence??
$endgroup$
– Math_QED
Jan 2 at 19:34
$begingroup$
Are you calculating the limit of that sequence or the series of the sequence??
$endgroup$
– Math_QED
Jan 2 at 19:34
$begingroup$
Regarding the WA link, note that you write x!/x^x, but sum over j...
$endgroup$
– Wojowu
Jan 2 at 19:35
$begingroup$
Regarding the WA link, note that you write x!/x^x, but sum over j...
$endgroup$
– Wojowu
Jan 2 at 19:35
$begingroup$
The reason that wolfram alpha says your sum diverges is as Wojowu says you used the variable $x$ instead of $j$ in that sum. On top of this you need to start the index at $j=1$ on wolfram because of its convention with $0^0$. you can see this here: wolframalpha.com/input/?i=sum+j!%2Fj%5Ej,+j%3D1..infinity Contrary to the answer youve accepted the sum actually does converge, and you can show this pretty easily with a comparison with $sum frac{2}{n^2}$
$endgroup$
– B.Martin
Jan 3 at 0:19
$begingroup$
The reason that wolfram alpha says your sum diverges is as Wojowu says you used the variable $x$ instead of $j$ in that sum. On top of this you need to start the index at $j=1$ on wolfram because of its convention with $0^0$. you can see this here: wolframalpha.com/input/?i=sum+j!%2Fj%5Ej,+j%3D1..infinity Contrary to the answer youve accepted the sum actually does converge, and you can show this pretty easily with a comparison with $sum frac{2}{n^2}$
$endgroup$
– B.Martin
Jan 3 at 0:19
add a comment |
1 Answer
1
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oldest
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Split it in terms of the form $frac{n}{x}$, with $n$ varying from $1$ to $x$. Clearly this is bounded above by the first term, which is $frac{1}{x}$, which is null, so the sequence tends to $0$.
And why does Wolfram Alpha disagree with me when I do the infinite summation of x!/x^x and it says by the limit test that the summation diverges:
Because, while the sequence $$left(dfrac{x!}{x^x}right)$$ converges, the series $${sum}dfrac{x!}{x^x}$$ does not.
answered Jan 2 at 19:37
user3482749user3482749
4,296919
$endgroup$
$begingroup$
But according to Khan Academy here: khanacademy.org/math/ap-calculus-bc/bc-series-new/modal/v/…, the improper integral of a function is an upper bound for its infinite summation, proving a very good argument for its convergence. On Wolfram, when I integrate x!/x^x from 0 to infinity I get something around 2.51791766302214. Shouldn't that be an upper bound for the series' summation, proving its convergence?
$endgroup$
– Math Bob
Jan 2 at 20:48
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Split it in terms of the form $frac{n}{x}$, with $n$ varying from $1$ to $x$. Clearly this is bounded above by the first term, which is $frac{1}{x}$, which is null, so the sequence tends to $0$.
And why does Wolfram Alpha disagree with me when I do the infinite summation of x!/x^x and it says by the limit test that the summation diverges:
Because, while the sequence $$left(dfrac{x!}{x^x}right)$$ converges, the series $${sum}dfrac{x!}{x^x}$$ does not.
answered Jan 2 at 19:37
user3482749user3482749
4,296919
$endgroup$
$begingroup$
But according to Khan Academy here: khanacademy.org/math/ap-calculus-bc/bc-series-new/modal/v/…, the improper integral of a function is an upper bound for its infinite summation, proving a very good argument for its convergence. On Wolfram, when I integrate x!/x^x from 0 to infinity I get something around 2.51791766302214. Shouldn't that be an upper bound for the series' summation, proving its convergence?
$endgroup$
– Math Bob
Jan 2 at 20:48
add a comment |
$begingroup$
Split it in terms of the form $frac{n}{x}$, with $n$ varying from $1$ to $x$. Clearly this is bounded above by the first term, which is $frac{1}{x}$, which is null, so the sequence tends to $0$.
And why does Wolfram Alpha disagree with me when I do the infinite summation of x!/x^x and it says by the limit test that the summation diverges:
Because, while the sequence $$left(dfrac{x!}{x^x}right)$$ converges, the series $${sum}dfrac{x!}{x^x}$$ does not.
answered Jan 2 at 19:37
user3482749user3482749
4,296919
$endgroup$
$begingroup$
But according to Khan Academy here: khanacademy.org/math/ap-calculus-bc/bc-series-new/modal/v/…, the improper integral of a function is an upper bound for its infinite summation, proving a very good argument for its convergence. On Wolfram, when I integrate x!/x^x from 0 to infinity I get something around 2.51791766302214. Shouldn't that be an upper bound for the series' summation, proving its convergence?
$endgroup$
– Math Bob
Jan 2 at 20:48
add a comment |
$begingroup$
Split it in terms of the form $frac{n}{x}$, with $n$ varying from $1$ to $x$. Clearly this is bounded above by the first term, which is $frac{1}{x}$, which is null, so the sequence tends to $0$.
And why does Wolfram Alpha disagree with me when I do the infinite summation of x!/x^x and it says by the limit test that the summation diverges:
Because, while the sequence $$left(dfrac{x!}{x^x}right)$$ converges, the series $${sum}dfrac{x!}{x^x}$$ does not.
answered Jan 2 at 19:37
user3482749user3482749
4,296919
$endgroup$
Split it in terms of the form $frac{n}{x}$, with $n$ varying from $1$ to $x$. Clearly this is bounded above by the first term, which is $frac{1}{x}$, which is null, so the sequence tends to $0$.
And why does Wolfram Alpha disagree with me when I do the infinite summation of x!/x^x and it says by the limit test that the summation diverges:
Because, while the sequence $$left(dfrac{x!}{x^x}right)$$ converges, the series $${sum}dfrac{x!}{x^x}$$ does not.
answered Jan 2 at 19:37
user3482749user3482749
4,296919
answered Jan 2 at 19:37
user3482749user3482749
4,296919
answered Jan 2 at 19:37
user3482749user3482749
4,296919
answered Jan 2 at 19:37
user3482749user3482749
4,296919
4,296919
$begingroup$
But according to Khan Academy here: khanacademy.org/math/ap-calculus-bc/bc-series-new/modal/v/…, the improper integral of a function is an upper bound for its infinite summation, proving a very good argument for its convergence. On Wolfram, when I integrate x!/x^x from 0 to infinity I get something around 2.51791766302214. Shouldn't that be an upper bound for the series' summation, proving its convergence?
$endgroup$
– Math Bob
Jan 2 at 20:48
add a comment |
$begingroup$
But according to Khan Academy here: khanacademy.org/math/ap-calculus-bc/bc-series-new/modal/v/…, the improper integral of a function is an upper bound for its infinite summation, proving a very good argument for its convergence. On Wolfram, when I integrate x!/x^x from 0 to infinity I get something around 2.51791766302214. Shouldn't that be an upper bound for the series' summation, proving its convergence?
$endgroup$
– Math Bob
Jan 2 at 20:48
$begingroup$
But according to Khan Academy here: khanacademy.org/math/ap-calculus-bc/bc-series-new/modal/v/…, the improper integral of a function is an upper bound for its infinite summation, proving a very good argument for its convergence. On Wolfram, when I integrate x!/x^x from 0 to infinity I get something around 2.51791766302214. Shouldn't that be an upper bound for the series' summation, proving its convergence?
$endgroup$
– Math Bob
Jan 2 at 20:48
$begingroup$
But according to Khan Academy here: khanacademy.org/math/ap-calculus-bc/bc-series-new/modal/v/…, the improper integral of a function is an upper bound for its infinite summation, proving a very good argument for its convergence. On Wolfram, when I integrate x!/x^x from 0 to infinity I get something around 2.51791766302214. Shouldn't that be an upper bound for the series' summation, proving its convergence?
$endgroup$
– Math Bob
Jan 2 at 20:48
add a comment |
1
$begingroup$
The limit of a sequence and the sum of a series are different things ...
$endgroup$
– Martin R
Jan 2 at 19:34
$begingroup$
Are you calculating the limit of that sequence or the series of the sequence??
$endgroup$
– Math_QED
Jan 2 at 19:34
$begingroup$
Regarding the WA link, note that you write x!/x^x, but sum over j...
$endgroup$
– Wojowu
Jan 2 at 19:35
$begingroup$
The reason that wolfram alpha says your sum diverges is as Wojowu says you used the variable $x$ instead of $j$ in that sum. On top of this you need to start the index at $j=1$ on wolfram because of its convention with $0^0$. you can see this here: wolframalpha.com/input/?i=sum+j!%2Fj%5Ej,+j%3D1..infinity Contrary to the answer youve accepted the sum actually does converge, and you can show this pretty easily with a comparison with $sum frac{2}{n^2}$
$endgroup$
– B.Martin
Jan 3 at 0:19