The limit of a sequence $u_n+1=exp(u_n)+u_n$ [closed]

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Please help me to find the limit of this sequence :

$ u_0=-2017;$ and $; u_n+1 =e^u_n+u_n $

I don't know if the limit exist or not.

sequences-and-series recurrence-relations

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edited Jul 18 at 20:07

amWhy

189k25219431

asked Jul 18 at 20:02

user574907

236

closed as off-topic by amWhy, José Carlos Santos, Shailesh, Alan Wang, Parcly Taxel Jul 19 at 1:10

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, Shailesh, Alan Wang, Parcly Taxel

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

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  HINT: $$u_n+1-u_n = e^u_n>0$$ implies that the sequence is strictly increasing. Thus the limit necessarily exists (possibly $+ infty$). Now, $$u_n+1-u_0= sum_k=0^n e^u_k > sum_k=0^n e^u_0 = u_0 cdot (n+1) to infty$$
  – Crostul
  Jul 18 at 20:08
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Hmm... raises the question of whether it's "converges to infinity" or "diverges to infinity".
  – Robert Wolfe
  Jul 18 at 20:09
  

  
  
  
  

add a comment | 

up vote
0
down vote

favorite

Please help me to find the limit of this sequence :

$ u_0=-2017;$ and $; u_n+1 =e^u_n+u_n $

I don't know if the limit exist or not.

sequences-and-series recurrence-relations

share|cite|improve this question

edited Jul 18 at 20:07

amWhy

189k25219431

asked Jul 18 at 20:02

user574907

236

closed as off-topic by amWhy, José Carlos Santos, Shailesh, Alan Wang, Parcly Taxel Jul 19 at 1:10

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, Shailesh, Alan Wang, Parcly Taxel

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

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  HINT: $$u_n+1-u_n = e^u_n>0$$ implies that the sequence is strictly increasing. Thus the limit necessarily exists (possibly $+ infty$). Now, $$u_n+1-u_0= sum_k=0^n e^u_k > sum_k=0^n e^u_0 = u_0 cdot (n+1) to infty$$
  – Crostul
  Jul 18 at 20:08
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Hmm... raises the question of whether it's "converges to infinity" or "diverges to infinity".
  – Robert Wolfe
  Jul 18 at 20:09
  

  
  
  
  

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

Please help me to find the limit of this sequence :

$ u_0=-2017;$ and $; u_n+1 =e^u_n+u_n $

I don't know if the limit exist or not.

sequences-and-series recurrence-relations

share|cite|improve this question

edited Jul 18 at 20:07

amWhy

189k25219431

asked Jul 18 at 20:02

user574907

236

Please help me to find the limit of this sequence :

$ u_0=-2017;$ and $; u_n+1 =e^u_n+u_n $

I don't know if the limit exist or not.

sequences-and-series recurrence-relations

share|cite|improve this question

edited Jul 18 at 20:07

amWhy

189k25219431

asked Jul 18 at 20:02

user574907

236

share|cite|improve this question

share|cite|improve this question

edited Jul 18 at 20:07

amWhy

189k25219431

edited Jul 18 at 20:07

amWhy

189k25219431

edited Jul 18 at 20:07

amWhy

189k25219431

189k25219431

asked Jul 18 at 20:02

user574907

236

asked Jul 18 at 20:02

user574907

236

asked Jul 18 at 20:02

user574907

236

236

closed as off-topic by amWhy, José Carlos Santos, Shailesh, Alan Wang, Parcly Taxel Jul 19 at 1:10

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, Shailesh, Alan Wang, Parcly Taxel

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

closed as off-topic by amWhy, José Carlos Santos, Shailesh, Alan Wang, Parcly Taxel Jul 19 at 1:10

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, Shailesh, Alan Wang, Parcly Taxel

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

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  HINT: $$u_n+1-u_n = e^u_n>0$$ implies that the sequence is strictly increasing. Thus the limit necessarily exists (possibly $+ infty$). Now, $$u_n+1-u_0= sum_k=0^n e^u_k > sum_k=0^n e^u_0 = u_0 cdot (n+1) to infty$$
  – Crostul
  Jul 18 at 20:08
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Hmm... raises the question of whether it's "converges to infinity" or "diverges to infinity".
  – Robert Wolfe
  Jul 18 at 20:09
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  HINT: $$u_n+1-u_n = e^u_n>0$$ implies that the sequence is strictly increasing. Thus the limit necessarily exists (possibly $+ infty$). Now, $$u_n+1-u_0= sum_k=0^n e^u_k > sum_k=0^n e^u_0 = u_0 cdot (n+1) to infty$$
  – Crostul
  Jul 18 at 20:08
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Hmm... raises the question of whether it's "converges to infinity" or "diverges to infinity".
  – Robert Wolfe
  Jul 18 at 20:09
  

  
  
  
  

1

1

HINT: $$u_n+1-u_n = e^u_n>0$$ implies that the sequence is strictly increasing. Thus the limit necessarily exists (possibly $+ infty$). Now, $$u_n+1-u_0= sum_k=0^n e^u_k > sum_k=0^n e^u_0 = u_0 cdot (n+1) to infty$$
– Crostul
Jul 18 at 20:08

HINT: $$u_n+1-u_n = e^u_n>0$$ implies that the sequence is strictly increasing. Thus the limit necessarily exists (possibly $+ infty$). Now, $$u_n+1-u_0= sum_k=0^n e^u_k > sum_k=0^n e^u_0 = u_0 cdot (n+1) to infty$$
– Crostul
Jul 18 at 20:08

Hmm... raises the question of whether it's "converges to infinity" or "diverges to infinity".
– Robert Wolfe
Jul 18 at 20:09

Hmm... raises the question of whether it's "converges to infinity" or "diverges to infinity".
– Robert Wolfe
Jul 18 at 20:09

add a comment | 

1 Answer
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accepted

If the limit $l$ exists, $l=e^l+l$ implies $e^l=0$ impossible.

share|cite|improve this answer

answered Jul 18 at 20:08

Tsemo Aristide

51.2k11243

add a comment | 

1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
10
down vote



accepted

If the limit $l$ exists, $l=e^l+l$ implies $e^l=0$ impossible.

share|cite|improve this answer

answered Jul 18 at 20:08

Tsemo Aristide

51.2k11243

add a comment | 

up vote
10
down vote



accepted

If the limit $l$ exists, $l=e^l+l$ implies $e^l=0$ impossible.

share|cite|improve this answer

answered Jul 18 at 20:08

Tsemo Aristide

51.2k11243

add a comment | 

up vote
10
down vote



accepted

up vote
10
down vote



accepted

If the limit $l$ exists, $l=e^l+l$ implies $e^l=0$ impossible.

share|cite|improve this answer

answered Jul 18 at 20:08

Tsemo Aristide

51.2k11243

If the limit $l$ exists, $l=e^l+l$ implies $e^l=0$ impossible.

share|cite|improve this answer

answered Jul 18 at 20:08

Tsemo Aristide

51.2k11243

share|cite|improve this answer

share|cite|improve this answer

answered Jul 18 at 20:08

Tsemo Aristide

51.2k11243

answered Jul 18 at 20:08

Tsemo Aristide

51.2k11243

answered Jul 18 at 20:08

Tsemo Aristide

51.2k11243

51.2k11243

add a comment | 

add a comment |