Infinite sum: $sum_x=1^inftyfracln xe^x$ [closed]

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I have tried to find this infinite sum all over the place but could not find it. Does anyone know a method for finding this sum:
$$lim_n to infty
sum_x=1^n ln(x)over e^x$$

Any help is greatly appreciated👍🏻

calculus real-analysis summation

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edited Jul 25 at 14:52

saulspatz

10.4k21323

asked Jul 25 at 14:43

Josh Messing

113

closed as off-topic by amWhy, Nosrati, José Carlos Santos, Chris Custer, max_zorn Jul 26 at 0:04

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." – amWhy, José Carlos Santos, Chris Custer, max_zorn

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  9
  

  
  
  
  
  
  

  
  Is there any reason for suspecting a closed form exists?
  – egreg
  Jul 25 at 14:47
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Are you looking for a closed form, or do you want to evaluate the sum numerically?
  – saulspatz
  Jul 25 at 14:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  What do you mean with "over" in your code? Do you mean $binomln(x)e^x$ or $ln(x)^e^x$
  – Cornman
  Jul 25 at 14:54
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Cornman He means "divided by."
  – saulspatz
  Jul 25 at 14:56
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  That is a good point egreg. I suppose I would be just as satisfied if I could be some how shown that there cannot be any closed form solution. If there is a closed form solution I would be looking for a method to find the closed form solution exactly. Not a numerical approximation
  – Josh Messing
  Jul 25 at 14:59
  
  

  
  
  
  

 | 
show 5 more comments

up vote
1
down vote

favorite

I have tried to find this infinite sum all over the place but could not find it. Does anyone know a method for finding this sum:
$$lim_n to infty
sum_x=1^n ln(x)over e^x$$

Any help is greatly appreciated👍🏻

calculus real-analysis summation

share|cite|improve this question

edited Jul 25 at 14:52

saulspatz

10.4k21323

asked Jul 25 at 14:43

Josh Messing

113

closed as off-topic by amWhy, Nosrati, José Carlos Santos, Chris Custer, max_zorn Jul 26 at 0:04

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." – amWhy, José Carlos Santos, Chris Custer, max_zorn

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  9
  

  
  
  
  
  
  

  
  Is there any reason for suspecting a closed form exists?
  – egreg
  Jul 25 at 14:47
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Are you looking for a closed form, or do you want to evaluate the sum numerically?
  – saulspatz
  Jul 25 at 14:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  What do you mean with "over" in your code? Do you mean $binomln(x)e^x$ or $ln(x)^e^x$
  – Cornman
  Jul 25 at 14:54
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Cornman He means "divided by."
  – saulspatz
  Jul 25 at 14:56
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  That is a good point egreg. I suppose I would be just as satisfied if I could be some how shown that there cannot be any closed form solution. If there is a closed form solution I would be looking for a method to find the closed form solution exactly. Not a numerical approximation
  – Josh Messing
  Jul 25 at 14:59
  
  

  
  
  
  

 | 
show 5 more comments

up vote
1
down vote

favorite

up vote
1
down vote

favorite

I have tried to find this infinite sum all over the place but could not find it. Does anyone know a method for finding this sum:
$$lim_n to infty
sum_x=1^n ln(x)over e^x$$

Any help is greatly appreciated👍🏻

calculus real-analysis summation

share|cite|improve this question

edited Jul 25 at 14:52

saulspatz

10.4k21323

asked Jul 25 at 14:43

Josh Messing

113

I have tried to find this infinite sum all over the place but could not find it. Does anyone know a method for finding this sum:
$$lim_n to infty
sum_x=1^n ln(x)over e^x$$

Any help is greatly appreciated👍🏻

calculus real-analysis summation

share|cite|improve this question

edited Jul 25 at 14:52

saulspatz

10.4k21323

asked Jul 25 at 14:43

Josh Messing

113

share|cite|improve this question

share|cite|improve this question

edited Jul 25 at 14:52

saulspatz

10.4k21323

edited Jul 25 at 14:52

saulspatz

10.4k21323

edited Jul 25 at 14:52

saulspatz

10.4k21323

10.4k21323

asked Jul 25 at 14:43

Josh Messing

113

asked Jul 25 at 14:43

Josh Messing

113

asked Jul 25 at 14:43

Josh Messing

113

113

closed as off-topic by amWhy, Nosrati, José Carlos Santos, Chris Custer, max_zorn Jul 26 at 0:04

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." – amWhy, José Carlos Santos, Chris Custer, max_zorn

If this question can be reworded to fit the rules in the help center, please edit the question.

closed as off-topic by amWhy, Nosrati, José Carlos Santos, Chris Custer, max_zorn Jul 26 at 0:04

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." – amWhy, José Carlos Santos, Chris Custer, max_zorn

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  9
  

  
  
  
  
  
  

  
  Is there any reason for suspecting a closed form exists?
  – egreg
  Jul 25 at 14:47
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Are you looking for a closed form, or do you want to evaluate the sum numerically?
  – saulspatz
  Jul 25 at 14:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  What do you mean with "over" in your code? Do you mean $binomln(x)e^x$ or $ln(x)^e^x$
  – Cornman
  Jul 25 at 14:54
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Cornman He means "divided by."
  – saulspatz
  Jul 25 at 14:56
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  That is a good point egreg. I suppose I would be just as satisfied if I could be some how shown that there cannot be any closed form solution. If there is a closed form solution I would be looking for a method to find the closed form solution exactly. Not a numerical approximation
  – Josh Messing
  Jul 25 at 14:59
  
  

  
  
  
  

 | 
show 5 more comments

• 
  
  
  

  
  9
  

  
  
  
  
  
  

  
  Is there any reason for suspecting a closed form exists?
  – egreg
  Jul 25 at 14:47
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Are you looking for a closed form, or do you want to evaluate the sum numerically?
  – saulspatz
  Jul 25 at 14:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  What do you mean with "over" in your code? Do you mean $binomln(x)e^x$ or $ln(x)^e^x$
  – Cornman
  Jul 25 at 14:54
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Cornman He means "divided by."
  – saulspatz
  Jul 25 at 14:56
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  That is a good point egreg. I suppose I would be just as satisfied if I could be some how shown that there cannot be any closed form solution. If there is a closed form solution I would be looking for a method to find the closed form solution exactly. Not a numerical approximation
  – Josh Messing
  Jul 25 at 14:59
  
  

  
  
  
  

9

9

Is there any reason for suspecting a closed form exists?
– egreg
Jul 25 at 14:47

Is there any reason for suspecting a closed form exists?
– egreg
Jul 25 at 14:47

1

1

Are you looking for a closed form, or do you want to evaluate the sum numerically?
– saulspatz
Jul 25 at 14:53

Are you looking for a closed form, or do you want to evaluate the sum numerically?
– saulspatz
Jul 25 at 14:53

What do you mean with "over" in your code? Do you mean $binomln(x)e^x$ or $ln(x)^e^x$
– Cornman
Jul 25 at 14:54

What do you mean with "over" in your code? Do you mean $binomln(x)e^x$ or $ln(x)^e^x$
– Cornman
Jul 25 at 14:54

@Cornman He means "divided by."
– saulspatz
Jul 25 at 14:56

@Cornman He means "divided by."
– saulspatz
Jul 25 at 14:56

That is a good point egreg. I suppose I would be just as satisfied if I could be some how shown that there cannot be any closed form solution. If there is a closed form solution I would be looking for a method to find the closed form solution exactly. Not a numerical approximation
– Josh Messing
Jul 25 at 14:59

That is a good point egreg. I suppose I would be just as satisfied if I could be some how shown that there cannot be any closed form solution. If there is a closed form solution I would be looking for a method to find the closed form solution exactly. Not a numerical approximation
– Josh Messing
Jul 25 at 14:59

 | 
show 5 more comments

1 Answer
1

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up vote
4
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$$S=sum_k=1^infty fracln(k)e^k$$
It is easy to prove the convergence. That is not the question.

The question is about the existence, or not, of a closed form. In other words, if $S$ is related to a standard special function.

Consider the kind of special functions called "Polylogarithm" :
$$textLi_nu(z)=sum_k=1^infty fracz^kk^nu=sum_k=1^infty e^-nu,ln(k)z^k$$
$$fracpartial partial nutextLi_nu(z)=-sum_k=1^infty ln(k)e^-nu,ln(k)z^k=-sum_k=1^infty fracln(k)k^nuz^k$$
In the particular case
$quadnu=0quadtextandquad z=frac1equad$ this leads to
$quad fracpartial partial nutextLi_nu(z)=-sum_k=1^infty fracln(k)e^k$
$$sum_k=1^infty fracln(k)e^k=-left(fracpartial partial nutextLi_nu(z)right)left(nu=0:,:z=1/eright)$$
This special function is implemented in WolframAlpha as $textPolyLog^(1,,,0)(nu,z)$. The first exponent means the partial derivative with regard to $nu$. The second exponent means the partial derivative with regard to $z$, which in the present case is of degree $0$, that is the function itself.

That is why WolframAlpha gives the result on the form :
$$sum_k=1^infty fracln(k)e^k=-textPolyLog^(1,,,0)left(0,frac1eright)simeq 0.1920928...$$

share|cite|improve this answer

answered Jul 25 at 16:17

JJacquelin

39.9k21649

add a comment | 

1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
4
down vote

$$S=sum_k=1^infty fracln(k)e^k$$
It is easy to prove the convergence. That is not the question.

The question is about the existence, or not, of a closed form. In other words, if $S$ is related to a standard special function.

Consider the kind of special functions called "Polylogarithm" :
$$textLi_nu(z)=sum_k=1^infty fracz^kk^nu=sum_k=1^infty e^-nu,ln(k)z^k$$
$$fracpartial partial nutextLi_nu(z)=-sum_k=1^infty ln(k)e^-nu,ln(k)z^k=-sum_k=1^infty fracln(k)k^nuz^k$$
In the particular case
$quadnu=0quadtextandquad z=frac1equad$ this leads to
$quad fracpartial partial nutextLi_nu(z)=-sum_k=1^infty fracln(k)e^k$
$$sum_k=1^infty fracln(k)e^k=-left(fracpartial partial nutextLi_nu(z)right)left(nu=0:,:z=1/eright)$$
This special function is implemented in WolframAlpha as $textPolyLog^(1,,,0)(nu,z)$. The first exponent means the partial derivative with regard to $nu$. The second exponent means the partial derivative with regard to $z$, which in the present case is of degree $0$, that is the function itself.

That is why WolframAlpha gives the result on the form :
$$sum_k=1^infty fracln(k)e^k=-textPolyLog^(1,,,0)left(0,frac1eright)simeq 0.1920928...$$

share|cite|improve this answer

answered Jul 25 at 16:17

JJacquelin

39.9k21649

add a comment | 

up vote
4
down vote

$$S=sum_k=1^infty fracln(k)e^k$$
It is easy to prove the convergence. That is not the question.

The question is about the existence, or not, of a closed form. In other words, if $S$ is related to a standard special function.

Consider the kind of special functions called "Polylogarithm" :
$$textLi_nu(z)=sum_k=1^infty fracz^kk^nu=sum_k=1^infty e^-nu,ln(k)z^k$$
$$fracpartial partial nutextLi_nu(z)=-sum_k=1^infty ln(k)e^-nu,ln(k)z^k=-sum_k=1^infty fracln(k)k^nuz^k$$
In the particular case
$quadnu=0quadtextandquad z=frac1equad$ this leads to
$quad fracpartial partial nutextLi_nu(z)=-sum_k=1^infty fracln(k)e^k$
$$sum_k=1^infty fracln(k)e^k=-left(fracpartial partial nutextLi_nu(z)right)left(nu=0:,:z=1/eright)$$
This special function is implemented in WolframAlpha as $textPolyLog^(1,,,0)(nu,z)$. The first exponent means the partial derivative with regard to $nu$. The second exponent means the partial derivative with regard to $z$, which in the present case is of degree $0$, that is the function itself.

That is why WolframAlpha gives the result on the form :
$$sum_k=1^infty fracln(k)e^k=-textPolyLog^(1,,,0)left(0,frac1eright)simeq 0.1920928...$$

share|cite|improve this answer

answered Jul 25 at 16:17

JJacquelin

39.9k21649

add a comment | 

up vote
4
down vote

up vote
4
down vote

$$S=sum_k=1^infty fracln(k)e^k$$
It is easy to prove the convergence. That is not the question.

The question is about the existence, or not, of a closed form. In other words, if $S$ is related to a standard special function.

Consider the kind of special functions called "Polylogarithm" :
$$textLi_nu(z)=sum_k=1^infty fracz^kk^nu=sum_k=1^infty e^-nu,ln(k)z^k$$
$$fracpartial partial nutextLi_nu(z)=-sum_k=1^infty ln(k)e^-nu,ln(k)z^k=-sum_k=1^infty fracln(k)k^nuz^k$$
In the particular case
$quadnu=0quadtextandquad z=frac1equad$ this leads to
$quad fracpartial partial nutextLi_nu(z)=-sum_k=1^infty fracln(k)e^k$
$$sum_k=1^infty fracln(k)e^k=-left(fracpartial partial nutextLi_nu(z)right)left(nu=0:,:z=1/eright)$$
This special function is implemented in WolframAlpha as $textPolyLog^(1,,,0)(nu,z)$. The first exponent means the partial derivative with regard to $nu$. The second exponent means the partial derivative with regard to $z$, which in the present case is of degree $0$, that is the function itself.

That is why WolframAlpha gives the result on the form :
$$sum_k=1^infty fracln(k)e^k=-textPolyLog^(1,,,0)left(0,frac1eright)simeq 0.1920928...$$

share|cite|improve this answer

answered Jul 25 at 16:17

JJacquelin

39.9k21649

$$S=sum_k=1^infty fracln(k)e^k$$
It is easy to prove the convergence. That is not the question.

The question is about the existence, or not, of a closed form. In other words, if $S$ is related to a standard special function.

Consider the kind of special functions called "Polylogarithm" :
$$textLi_nu(z)=sum_k=1^infty fracz^kk^nu=sum_k=1^infty e^-nu,ln(k)z^k$$
$$fracpartial partial nutextLi_nu(z)=-sum_k=1^infty ln(k)e^-nu,ln(k)z^k=-sum_k=1^infty fracln(k)k^nuz^k$$
In the particular case
$quadnu=0quadtextandquad z=frac1equad$ this leads to
$quad fracpartial partial nutextLi_nu(z)=-sum_k=1^infty fracln(k)e^k$
$$sum_k=1^infty fracln(k)e^k=-left(fracpartial partial nutextLi_nu(z)right)left(nu=0:,:z=1/eright)$$
This special function is implemented in WolframAlpha as $textPolyLog^(1,,,0)(nu,z)$. The first exponent means the partial derivative with regard to $nu$. The second exponent means the partial derivative with regard to $z$, which in the present case is of degree $0$, that is the function itself.

That is why WolframAlpha gives the result on the form :
$$sum_k=1^infty fracln(k)e^k=-textPolyLog^(1,,,0)left(0,frac1eright)simeq 0.1920928...$$

share|cite|improve this answer

answered Jul 25 at 16:17

JJacquelin

39.9k21649

share|cite|improve this answer

share|cite|improve this answer

answered Jul 25 at 16:17

JJacquelin

39.9k21649

answered Jul 25 at 16:17

JJacquelin

39.9k21649

answered Jul 25 at 16:17

JJacquelin

39.9k21649

39.9k21649

add a comment | 

add a comment |