Is $lim_{x to infty }sum_{n=1}^infty frac{n!}{e^{n^x}}=e^{-1}$? [closed]












0














Is it the case that:
$$ lim_{x to infty }sum_{n=1}^infty frac{n!}{e^{n^x}}=e^{-1}$$



And why?










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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.













  • desmos.com/calculator/zy8c7ucm4h
    – Mason
    Dec 1 '18 at 4:23












  • You know... Maybe write it as a taylor series and then take $x$ large is the right approach...
    – Mason
    Dec 1 '18 at 4:28






  • 1




    When $n=1$, you have $1/e$.
    – i707107
    Dec 1 '18 at 4:33






  • 1




    You may prove the convergence of other terms to zero as $xrightarrowinfty$.
    – i707107
    Dec 1 '18 at 4:34










  • Oh! Makes sense. Thanks.
    – Mason
    Dec 1 '18 at 4:34
















0














Is it the case that:
$$ lim_{x to infty }sum_{n=1}^infty frac{n!}{e^{n^x}}=e^{-1}$$



And why?










share|cite|improve this 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.













  • desmos.com/calculator/zy8c7ucm4h
    – Mason
    Dec 1 '18 at 4:23












  • You know... Maybe write it as a taylor series and then take $x$ large is the right approach...
    – Mason
    Dec 1 '18 at 4:28






  • 1




    When $n=1$, you have $1/e$.
    – i707107
    Dec 1 '18 at 4:33






  • 1




    You may prove the convergence of other terms to zero as $xrightarrowinfty$.
    – i707107
    Dec 1 '18 at 4:34










  • Oh! Makes sense. Thanks.
    – Mason
    Dec 1 '18 at 4:34














0












0








0







Is it the case that:
$$ lim_{x to infty }sum_{n=1}^infty frac{n!}{e^{n^x}}=e^{-1}$$



And why?










share|cite|improve this question















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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 1 '18 at 6:48









Martin Sleziak

44.7k8115271




44.7k8115271










asked Dec 1 '18 at 4:23









MasonMason

1,9591530




1,9591530




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.












  • desmos.com/calculator/zy8c7ucm4h
    – Mason
    Dec 1 '18 at 4:23












  • You know... Maybe write it as a taylor series and then take $x$ large is the right approach...
    – Mason
    Dec 1 '18 at 4:28






  • 1




    When $n=1$, you have $1/e$.
    – i707107
    Dec 1 '18 at 4:33






  • 1




    You may prove the convergence of other terms to zero as $xrightarrowinfty$.
    – i707107
    Dec 1 '18 at 4:34










  • Oh! Makes sense. Thanks.
    – Mason
    Dec 1 '18 at 4:34


















  • desmos.com/calculator/zy8c7ucm4h
    – Mason
    Dec 1 '18 at 4:23












  • You know... Maybe write it as a taylor series and then take $x$ large is the right approach...
    – Mason
    Dec 1 '18 at 4:28






  • 1




    When $n=1$, you have $1/e$.
    – i707107
    Dec 1 '18 at 4:33






  • 1




    You may prove the convergence of other terms to zero as $xrightarrowinfty$.
    – i707107
    Dec 1 '18 at 4:34










  • Oh! Makes sense. Thanks.
    – Mason
    Dec 1 '18 at 4:34
















desmos.com/calculator/zy8c7ucm4h
– Mason
Dec 1 '18 at 4:23






desmos.com/calculator/zy8c7ucm4h
– Mason
Dec 1 '18 at 4:23














You know... Maybe write it as a taylor series and then take $x$ large is the right approach...
– Mason
Dec 1 '18 at 4:28




You know... Maybe write it as a taylor series and then take $x$ large is the right approach...
– Mason
Dec 1 '18 at 4:28




1




1




When $n=1$, you have $1/e$.
– i707107
Dec 1 '18 at 4:33




When $n=1$, you have $1/e$.
– i707107
Dec 1 '18 at 4:33




1




1




You may prove the convergence of other terms to zero as $xrightarrowinfty$.
– i707107
Dec 1 '18 at 4:34




You may prove the convergence of other terms to zero as $xrightarrowinfty$.
– i707107
Dec 1 '18 at 4:34












Oh! Makes sense. Thanks.
– Mason
Dec 1 '18 at 4:34




Oh! Makes sense. Thanks.
– Mason
Dec 1 '18 at 4:34










1 Answer
1






active

oldest

votes


















1














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.






share|cite|improve this answer

















  • 1




    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.
    – xbh
    Dec 1 '18 at 4:52










  • 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?
    – Mason
    Dec 1 '18 at 4:59












  • Yeah, that would work.
    – xbh
    Dec 1 '18 at 5:00


















1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1














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.






share|cite|improve this answer

















  • 1




    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.
    – xbh
    Dec 1 '18 at 4:52










  • 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?
    – Mason
    Dec 1 '18 at 4:59












  • Yeah, that would work.
    – xbh
    Dec 1 '18 at 5:00
















1














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.






share|cite|improve this answer

















  • 1




    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.
    – xbh
    Dec 1 '18 at 4:52










  • 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?
    – Mason
    Dec 1 '18 at 4:59












  • Yeah, that would work.
    – xbh
    Dec 1 '18 at 5:00














1












1








1






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.






share|cite|improve this answer












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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 1 '18 at 4:38









MasonMason

1,9591530




1,9591530








  • 1




    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.
    – xbh
    Dec 1 '18 at 4:52










  • 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?
    – Mason
    Dec 1 '18 at 4:59












  • Yeah, that would work.
    – xbh
    Dec 1 '18 at 5:00














  • 1




    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.
    – xbh
    Dec 1 '18 at 4:52










  • 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?
    – Mason
    Dec 1 '18 at 4:59












  • Yeah, that would work.
    – xbh
    Dec 1 '18 at 5:00








1




1




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.
– xbh
Dec 1 '18 at 4:52




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.
– xbh
Dec 1 '18 at 4:52












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?
– Mason
Dec 1 '18 at 4:59






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?
– Mason
Dec 1 '18 at 4:59














Yeah, that would work.
– xbh
Dec 1 '18 at 5:00




Yeah, that would work.
– xbh
Dec 1 '18 at 5:00



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