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Under what conditions can we swap a limit with an infinite product? [closed]

A limit with infinite productInterchange of infinite product and limitElementary Infinite Limit QuestionInfinite product of limit elements.What is an infinite limit?does the limit product rule applies to $0$? can I apply the product rule with $c to infty$?limit with summation and productLimit of an infinite product while x tends to $1^-$Under what conditions can one change the set that the limit variable ranges over?Limit of product of infinite sequences and infinite series

1

$begingroup$

When is the following true: $$lim_ntoinftyprod^infty_k=1a_nk = prod^infty_k=1lim_ntoinftya_nk$$

real-analysis limits

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asked Mar 29 at 16:26

Joshua FarrellJoshua Farrell

681220

$endgroup$

closed as off-topic by Saad, YiFan, Shailesh, Leucippus, Dr. Mathva Mar 30 at 7:54

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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, YiFan, Shailesh, Leucippus, Dr. Mathva

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

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Hint: This can be rewritten as $$lim_n to infty lim_mto infty prod_k=1^m a_nk = lim_m to infty prod_k=1^m lim_nto infty a_nk$$ So, you are swapping the first two limits, then you are swapping the limit and the product. So, at the very least, the product of the limits must be defined. But, the question is really more when is this true: $$lim_nto infty lim_m to infty prod_k=1^m a_nk = lim_m to infty lim_n to infty prod_k=1^m a_nk$$ You are probably looking at something like the Dominated Convergence Theorem.
  $endgroup$
  – InterstellarProbe
  Mar 29 at 16:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Assuming $a_nk > 0$ and the infinite products are nonzero, you can take logarithm of both sides and transform this to a question of interchanging limit with sum. The standard cases are monotone convergence and dominated convergence.
  $endgroup$
  – Robert Israel
  Mar 29 at 16:41
  

  
  
  
  

add a comment | 

1

$begingroup$

When is the following true: $$lim_ntoinftyprod^infty_k=1a_nk = prod^infty_k=1lim_ntoinftya_nk$$

real-analysis limits

share|cite|improve this question

asked Mar 29 at 16:26

Joshua FarrellJoshua Farrell

681220

$endgroup$

closed as off-topic by Saad, YiFan, Shailesh, Leucippus, Dr. Mathva Mar 30 at 7:54

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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, YiFan, Shailesh, Leucippus, Dr. Mathva

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

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Hint: This can be rewritten as $$lim_n to infty lim_mto infty prod_k=1^m a_nk = lim_m to infty prod_k=1^m lim_nto infty a_nk$$ So, you are swapping the first two limits, then you are swapping the limit and the product. So, at the very least, the product of the limits must be defined. But, the question is really more when is this true: $$lim_nto infty lim_m to infty prod_k=1^m a_nk = lim_m to infty lim_n to infty prod_k=1^m a_nk$$ You are probably looking at something like the Dominated Convergence Theorem.
  $endgroup$
  – InterstellarProbe
  Mar 29 at 16:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Assuming $a_nk > 0$ and the infinite products are nonzero, you can take logarithm of both sides and transform this to a question of interchanging limit with sum. The standard cases are monotone convergence and dominated convergence.
  $endgroup$
  – Robert Israel
  Mar 29 at 16:41
  

  
  
  
  

add a comment | 

$begingroup$

When is the following true: $$lim_ntoinftyprod^infty_k=1a_nk = prod^infty_k=1lim_ntoinftya_nk$$

real-analysis limits

share|cite|improve this question

asked Mar 29 at 16:26

Joshua FarrellJoshua Farrell

681220

$endgroup$

share|cite|improve this question

asked Mar 29 at 16:26

Joshua FarrellJoshua Farrell

681220

share|cite|improve this question

asked Mar 29 at 16:26

Joshua FarrellJoshua Farrell

681220

asked Mar 29 at 16:26

Joshua FarrellJoshua Farrell

681220

asked Mar 29 at 16:26

Joshua FarrellJoshua Farrell

681220

1 Answer
1

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oldest

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0

$begingroup$

You have
$$lim_ntoinftyprod^infty_k=1a_nk = prod^infty_k=1lim_ntoinftya_nk$$

If and only if

$$lim_ntoinftylog(prod^infty_k=1a_nk) = log(prod^infty_k=1lim_ntoinftya_nk)$$ (Warning see footnote below ¹)

Since $log$ takes product to sum by setting $b_nk=log(a_nk)$ the above is equivalent to whether
$$lim_ntoinftysum^infty_k=1b_nk = sum^infty_k=1lim_ntoinftyb_nk$$

Now you can use the Dominated convergence theorem which in this setting says:

If $|b_nk|leq c_k$ for a sequence $c_k$ for which $sum_k=1^infty c_k$ exists and $lim_nrightarrowinftyb_nk = b_k$ then $sum_k=1^infty b_k$ exists and
$$lim_ntoinftysum^infty_k=1b_nk = sum^infty_k=1b_k$$

Translated back to products, you need that $|a_nk|leq c_k$ for a sequence such that $prod_k=1^infty c_k$ exists then if $lim_nrightarrowinfty a_nk = a_k$ then $prod_k=1^infty a_k$ exists and
$$lim_ntoinftyprod^infty_k=1a_nk = prod^infty_k=1a_k$$

---

¹ This approach only works if the $a_nk$ are positive. Otherwise you can't take $log$.

share|cite|improve this answer

edited Mar 29 at 17:10

answered Mar 29 at 17:04

YankoYanko

8,3842830

$endgroup$

add a comment | 

0

$begingroup$

You have
$$lim_ntoinftyprod^infty_k=1a_nk = prod^infty_k=1lim_ntoinftya_nk$$

If and only if

$$lim_ntoinftylog(prod^infty_k=1a_nk) = log(prod^infty_k=1lim_ntoinftya_nk)$$ (Warning see footnote below ¹)

Since $log$ takes product to sum by setting $b_nk=log(a_nk)$ the above is equivalent to whether
$$lim_ntoinftysum^infty_k=1b_nk = sum^infty_k=1lim_ntoinftyb_nk$$

Now you can use the Dominated convergence theorem which in this setting says:

If $|b_nk|leq c_k$ for a sequence $c_k$ for which $sum_k=1^infty c_k$ exists and $lim_nrightarrowinftyb_nk = b_k$ then $sum_k=1^infty b_k$ exists and
$$lim_ntoinftysum^infty_k=1b_nk = sum^infty_k=1b_k$$

Translated back to products, you need that $|a_nk|leq c_k$ for a sequence such that $prod_k=1^infty c_k$ exists then if $lim_nrightarrowinfty a_nk = a_k$ then $prod_k=1^infty a_k$ exists and
$$lim_ntoinftyprod^infty_k=1a_nk = prod^infty_k=1a_k$$

---

¹ This approach only works if the $a_nk$ are positive. Otherwise you can't take $log$.

share|cite|improve this answer

edited Mar 29 at 17:10

answered Mar 29 at 17:04

YankoYanko

8,3842830

$endgroup$

add a comment | 

0

$begingroup$

You have
$$lim_ntoinftyprod^infty_k=1a_nk = prod^infty_k=1lim_ntoinftya_nk$$

If and only if

$$lim_ntoinftylog(prod^infty_k=1a_nk) = log(prod^infty_k=1lim_ntoinftya_nk)$$ (Warning see footnote below ¹)

Since $log$ takes product to sum by setting $b_nk=log(a_nk)$ the above is equivalent to whether
$$lim_ntoinftysum^infty_k=1b_nk = sum^infty_k=1lim_ntoinftyb_nk$$

Now you can use the Dominated convergence theorem which in this setting says:

If $|b_nk|leq c_k$ for a sequence $c_k$ for which $sum_k=1^infty c_k$ exists and $lim_nrightarrowinftyb_nk = b_k$ then $sum_k=1^infty b_k$ exists and
$$lim_ntoinftysum^infty_k=1b_nk = sum^infty_k=1b_k$$

Translated back to products, you need that $|a_nk|leq c_k$ for a sequence such that $prod_k=1^infty c_k$ exists then if $lim_nrightarrowinfty a_nk = a_k$ then $prod_k=1^infty a_k$ exists and
$$lim_ntoinftyprod^infty_k=1a_nk = prod^infty_k=1a_k$$

---

¹ This approach only works if the $a_nk$ are positive. Otherwise you can't take $log$.

share|cite|improve this answer

edited Mar 29 at 17:10

answered Mar 29 at 17:04

YankoYanko

8,3842830

$endgroup$

add a comment | 

$begingroup$

You have
$$lim_ntoinftyprod^infty_k=1a_nk = prod^infty_k=1lim_ntoinftya_nk$$

If and only if

$$lim_ntoinftylog(prod^infty_k=1a_nk) = log(prod^infty_k=1lim_ntoinftya_nk)$$ (Warning see footnote below ¹)

Since $log$ takes product to sum by setting $b_nk=log(a_nk)$ the above is equivalent to whether
$$lim_ntoinftysum^infty_k=1b_nk = sum^infty_k=1lim_ntoinftyb_nk$$

Now you can use the Dominated convergence theorem which in this setting says:

If $|b_nk|leq c_k$ for a sequence $c_k$ for which $sum_k=1^infty c_k$ exists and $lim_nrightarrowinftyb_nk = b_k$ then $sum_k=1^infty b_k$ exists and
$$lim_ntoinftysum^infty_k=1b_nk = sum^infty_k=1b_k$$

Translated back to products, you need that $|a_nk|leq c_k$ for a sequence such that $prod_k=1^infty c_k$ exists then if $lim_nrightarrowinfty a_nk = a_k$ then $prod_k=1^infty a_k$ exists and
$$lim_ntoinftyprod^infty_k=1a_nk = prod^infty_k=1a_k$$

---

¹ This approach only works if the $a_nk$ are positive. Otherwise you can't take $log$.

share|cite|improve this answer

edited Mar 29 at 17:10

answered Mar 29 at 17:04

YankoYanko

8,3842830

$endgroup$

share|cite|improve this answer

edited Mar 29 at 17:10

answered Mar 29 at 17:04

YankoYanko

8,3842830

answered Mar 29 at 17:04

YankoYanko

8,3842830

answered Mar 29 at 17:04

YankoYanko

8,3842830