Is $lim_{x to infty }sum_{n=1}^infty frac{n!}{e^{n^x}}=e^{-1}$? [closed]
0
$begingroup$
Is it the case that:
$$ lim_{x to infty }sum_{n=1}^infty frac{n!}{e^{n^x}}=e^{-1}$$
And why?
sequences-and-series limits exponential-function
edited Dec 1 '18 at 6:48
Martin Sleziak
44.7k8117272
asked Dec 1 '18 at 4:23
MasonMason
1,9571530
$endgroup$
closed as off-topic by Saad, José Carlos Santos, Cesareo, Brahadeesh, Vidyanshu Mishra Dec 1 '18 at 13:32
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Saad, José Carlos Santos, Cesareo, Brahadeesh, Vidyanshu Mishra
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
0
$begingroup$
Is it the case that:
$$ lim_{x to infty }sum_{n=1}^infty frac{n!}{e^{n^x}}=e^{-1}$$
And why?
sequences-and-series limits exponential-function
edited Dec 1 '18 at 6:48
Martin Sleziak
44.7k8117272
asked Dec 1 '18 at 4:23
MasonMason
1,9571530
$endgroup$
closed as off-topic by Saad, José Carlos Santos, Cesareo, Brahadeesh, Vidyanshu Mishra Dec 1 '18 at 13:32
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Saad, José Carlos Santos, Cesareo, Brahadeesh, Vidyanshu Mishra
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
desmos.com/calculator/zy8c7ucm4h
$endgroup$
– Mason
Dec 1 '18 at 4:23
$begingroup$
You know... Maybe write it as a taylor series and then take $x$ large is the right approach...
$endgroup$
– Mason
Dec 1 '18 at 4:28
1
$begingroup$
When $n=1$, you have $1/e$.
$endgroup$
– i707107
Dec 1 '18 at 4:33
1
$begingroup$
You may prove the convergence of other terms to zero as $xrightarrowinfty$.
$endgroup$
– i707107
Dec 1 '18 at 4:34
$begingroup$
Oh! Makes sense. Thanks.
$endgroup$
– Mason
Dec 1 '18 at 4:34
add a comment |
0
0
0
$begingroup$
Is it the case that:
$$ lim_{x to infty }sum_{n=1}^infty frac{n!}{e^{n^x}}=e^{-1}$$
And why?
sequences-and-series limits exponential-function
edited Dec 1 '18 at 6:48
Martin Sleziak
44.7k8117272
asked Dec 1 '18 at 4:23
MasonMason
1,9571530
$endgroup$
Is it the case that:
$$ lim_{x to infty }sum_{n=1}^infty frac{n!}{e^{n^x}}=e^{-1}$$
And why?
sequences-and-series limits exponential-function
sequences-and-series limits exponential-function
edited Dec 1 '18 at 6:48
Martin Sleziak
44.7k8117272
asked Dec 1 '18 at 4:23
MasonMason
1,9571530
edited Dec 1 '18 at 6:48
Martin Sleziak
44.7k8117272
asked Dec 1 '18 at 4:23
MasonMason
1,9571530
edited Dec 1 '18 at 6:48
Martin Sleziak
44.7k8117272
edited Dec 1 '18 at 6:48
Martin Sleziak
44.7k8117272
edited Dec 1 '18 at 6:48
Martin Sleziak
44.7k8117272
44.7k8117272
asked Dec 1 '18 at 4:23
MasonMason
1,9571530
asked Dec 1 '18 at 4:23
MasonMason
1,9571530
asked Dec 1 '18 at 4:23
MasonMason
1,9571530
1,9571530
closed as off-topic by Saad, José Carlos Santos, Cesareo, Brahadeesh, Vidyanshu Mishra Dec 1 '18 at 13:32
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Saad, José Carlos Santos, Cesareo, Brahadeesh, Vidyanshu Mishra
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Saad, José Carlos Santos, Cesareo, Brahadeesh, Vidyanshu Mishra Dec 1 '18 at 13:32
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Saad, José Carlos Santos, Cesareo, Brahadeesh, Vidyanshu Mishra
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
desmos.com/calculator/zy8c7ucm4h
$endgroup$
– Mason
Dec 1 '18 at 4:23
$begingroup$
You know... Maybe write it as a taylor series and then take $x$ large is the right approach...
$endgroup$
– Mason
Dec 1 '18 at 4:28
1
$begingroup$
When $n=1$, you have $1/e$.
$endgroup$
– i707107
Dec 1 '18 at 4:33
1
$begingroup$
You may prove the convergence of other terms to zero as $xrightarrowinfty$.
$endgroup$
– i707107
Dec 1 '18 at 4:34
$begingroup$
Oh! Makes sense. Thanks.
$endgroup$
– Mason
Dec 1 '18 at 4:34
add a comment |
$begingroup$
desmos.com/calculator/zy8c7ucm4h
$endgroup$
– Mason
Dec 1 '18 at 4:23
$begingroup$
You know... Maybe write it as a taylor series and then take $x$ large is the right approach...
$endgroup$
– Mason
Dec 1 '18 at 4:28
1
$begingroup$
When $n=1$, you have $1/e$.
$endgroup$
– i707107
Dec 1 '18 at 4:33
1
$begingroup$
You may prove the convergence of other terms to zero as $xrightarrowinfty$.
$endgroup$
– i707107
Dec 1 '18 at 4:34
$begingroup$
Oh! Makes sense. Thanks.
$endgroup$
– Mason
Dec 1 '18 at 4:34
$begingroup$
desmos.com/calculator/zy8c7ucm4h
$endgroup$
– Mason
Dec 1 '18 at 4:23
$begingroup$
desmos.com/calculator/zy8c7ucm4h
$endgroup$
– Mason
Dec 1 '18 at 4:23
$begingroup$
You know... Maybe write it as a taylor series and then take $x$ large is the right approach...
$endgroup$
– Mason
Dec 1 '18 at 4:28
$begingroup$
You know... Maybe write it as a taylor series and then take $x$ large is the right approach...
$endgroup$
– Mason
Dec 1 '18 at 4:28
1
1
$begingroup$
When $n=1$, you have $1/e$.
$endgroup$
– i707107
Dec 1 '18 at 4:33
$begingroup$
When $n=1$, you have $1/e$.
$endgroup$
– i707107
Dec 1 '18 at 4:33
1
1
$begingroup$
You may prove the convergence of other terms to zero as $xrightarrowinfty$.
$endgroup$
– i707107
Dec 1 '18 at 4:34
$begingroup$
You may prove the convergence of other terms to zero as $xrightarrowinfty$.
$endgroup$
– i707107
Dec 1 '18 at 4:34
$begingroup$
Oh! Makes sense. Thanks.
$endgroup$
– Mason
Dec 1 '18 at 4:34
$begingroup$
Oh! Makes sense. Thanks.
$endgroup$
– Mason
Dec 1 '18 at 4:34
add a comment |
1 Answer
1
active
oldest
votes
1
$begingroup$
Ok! I get it.
I thought I was looking at some great mystery but it's very simple.
As user i707107 mentions in the comments.
$$sum_{n=1}^inftyfrac{n!}{e^{n^x}} =frac{1}{e}+sum_{n=2}^infty frac{n!}{e^{n^x}}$$ And as we take $x$ large the denominator dwarfs the numerator and these terms (and the entire sum) limit to zero.
answered Dec 1 '18 at 4:38
MasonMason
1,9571530
$endgroup$
1
$begingroup$
However, $$ lim_{xto+infty} sum_2^{+infty} frac {n!}{mathrm e^{n^x}} = sum_2^{+infty} lim_{xto+infty} frac {n!}{mathrm e^{n^x}}$$ needs justification. Although this is not hard to do either.
$endgroup$
– xbh
Dec 1 '18 at 4:52
$begingroup$
My analysis is really rusty: something like: continuous functions uniformly converging to $f(x)=0 implies $ it's ok to move the limit inside the sum?
$endgroup$
– Mason
Dec 1 '18 at 4:59
$begingroup$
Yeah, that would work.
$endgroup$
– xbh
Dec 1 '18 at 5:00
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
1
$begingroup$
Ok! I get it.
I thought I was looking at some great mystery but it's very simple.
As user i707107 mentions in the comments.
$$sum_{n=1}^inftyfrac{n!}{e^{n^x}} =frac{1}{e}+sum_{n=2}^infty frac{n!}{e^{n^x}}$$ And as we take $x$ large the denominator dwarfs the numerator and these terms (and the entire sum) limit to zero.
answered Dec 1 '18 at 4:38
MasonMason
1,9571530
$endgroup$
1
$begingroup$
However, $$ lim_{xto+infty} sum_2^{+infty} frac {n!}{mathrm e^{n^x}} = sum_2^{+infty} lim_{xto+infty} frac {n!}{mathrm e^{n^x}}$$ needs justification. Although this is not hard to do either.
$endgroup$
– xbh
Dec 1 '18 at 4:52
$begingroup$
My analysis is really rusty: something like: continuous functions uniformly converging to $f(x)=0 implies $ it's ok to move the limit inside the sum?
$endgroup$
– Mason
Dec 1 '18 at 4:59
$begingroup$
Yeah, that would work.
$endgroup$
– xbh
Dec 1 '18 at 5:00
add a comment |
1
$begingroup$
Ok! I get it.
I thought I was looking at some great mystery but it's very simple.
As user i707107 mentions in the comments.
$$sum_{n=1}^inftyfrac{n!}{e^{n^x}} =frac{1}{e}+sum_{n=2}^infty frac{n!}{e^{n^x}}$$ And as we take $x$ large the denominator dwarfs the numerator and these terms (and the entire sum) limit to zero.
answered Dec 1 '18 at 4:38
MasonMason
1,9571530
$endgroup$
1
$begingroup$
However, $$ lim_{xto+infty} sum_2^{+infty} frac {n!}{mathrm e^{n^x}} = sum_2^{+infty} lim_{xto+infty} frac {n!}{mathrm e^{n^x}}$$ needs justification. Although this is not hard to do either.
$endgroup$
– xbh
Dec 1 '18 at 4:52
$begingroup$
My analysis is really rusty: something like: continuous functions uniformly converging to $f(x)=0 implies $ it's ok to move the limit inside the sum?
$endgroup$
– Mason
Dec 1 '18 at 4:59
$begingroup$
Yeah, that would work.
$endgroup$
– xbh
Dec 1 '18 at 5:00
add a comment |
1
1
1
$begingroup$
Ok! I get it.
I thought I was looking at some great mystery but it's very simple.
As user i707107 mentions in the comments.
$$sum_{n=1}^inftyfrac{n!}{e^{n^x}} =frac{1}{e}+sum_{n=2}^infty frac{n!}{e^{n^x}}$$ And as we take $x$ large the denominator dwarfs the numerator and these terms (and the entire sum) limit to zero.
answered Dec 1 '18 at 4:38
MasonMason
1,9571530
$endgroup$
Ok! I get it.
I thought I was looking at some great mystery but it's very simple.
As user i707107 mentions in the comments.
$$sum_{n=1}^inftyfrac{n!}{e^{n^x}} =frac{1}{e}+sum_{n=2}^infty frac{n!}{e^{n^x}}$$ And as we take $x$ large the denominator dwarfs the numerator and these terms (and the entire sum) limit to zero.
answered Dec 1 '18 at 4:38
MasonMason
1,9571530
answered Dec 1 '18 at 4:38
MasonMason
1,9571530
answered Dec 1 '18 at 4:38
MasonMason
1,9571530
answered Dec 1 '18 at 4:38
MasonMason
1,9571530
1,9571530
1
$begingroup$
However, $$ lim_{xto+infty} sum_2^{+infty} frac {n!}{mathrm e^{n^x}} = sum_2^{+infty} lim_{xto+infty} frac {n!}{mathrm e^{n^x}}$$ needs justification. Although this is not hard to do either.
$endgroup$
– xbh
Dec 1 '18 at 4:52
$begingroup$
My analysis is really rusty: something like: continuous functions uniformly converging to $f(x)=0 implies $ it's ok to move the limit inside the sum?
$endgroup$
– Mason
Dec 1 '18 at 4:59
$begingroup$
Yeah, that would work.
$endgroup$
– xbh
Dec 1 '18 at 5:00
add a comment |
1
$begingroup$
However, $$ lim_{xto+infty} sum_2^{+infty} frac {n!}{mathrm e^{n^x}} = sum_2^{+infty} lim_{xto+infty} frac {n!}{mathrm e^{n^x}}$$ needs justification. Although this is not hard to do either.
$endgroup$
– xbh
Dec 1 '18 at 4:52
$begingroup$
My analysis is really rusty: something like: continuous functions uniformly converging to $f(x)=0 implies $ it's ok to move the limit inside the sum?
$endgroup$
– Mason
Dec 1 '18 at 4:59
$begingroup$
Yeah, that would work.
$endgroup$
– xbh
Dec 1 '18 at 5:00
1
1
$begingroup$
However, $$ lim_{xto+infty} sum_2^{+infty} frac {n!}{mathrm e^{n^x}} = sum_2^{+infty} lim_{xto+infty} frac {n!}{mathrm e^{n^x}}$$ needs justification. Although this is not hard to do either.
$endgroup$
– xbh
Dec 1 '18 at 4:52
$begingroup$
However, $$ lim_{xto+infty} sum_2^{+infty} frac {n!}{mathrm e^{n^x}} = sum_2^{+infty} lim_{xto+infty} frac {n!}{mathrm e^{n^x}}$$ needs justification. Although this is not hard to do either.
$endgroup$
– xbh
Dec 1 '18 at 4:52
$begingroup$
My analysis is really rusty: something like: continuous functions uniformly converging to $f(x)=0 implies $ it's ok to move the limit inside the sum?
$endgroup$
– Mason
Dec 1 '18 at 4:59
$begingroup$
My analysis is really rusty: something like: continuous functions uniformly converging to $f(x)=0 implies $ it's ok to move the limit inside the sum?
$endgroup$
– Mason
Dec 1 '18 at 4:59
$begingroup$
Yeah, that would work.
$endgroup$
– xbh
Dec 1 '18 at 5:00
$begingroup$
Yeah, that would work.
$endgroup$
– xbh
Dec 1 '18 at 5:00
add a comment |
$begingroup$
desmos.com/calculator/zy8c7ucm4h
$endgroup$
– Mason
Dec 1 '18 at 4:23
$begingroup$
You know... Maybe write it as a taylor series and then take $x$ large is the right approach...
$endgroup$
– Mason
Dec 1 '18 at 4:28
1
$begingroup$
When $n=1$, you have $1/e$.
$endgroup$
– i707107
Dec 1 '18 at 4:33
1
$begingroup$
You may prove the convergence of other terms to zero as $xrightarrowinfty$.
$endgroup$
– i707107
Dec 1 '18 at 4:34
$begingroup$
Oh! Makes sense. Thanks.
$endgroup$
– Mason
Dec 1 '18 at 4:34