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Trouble with evaluating the limit of a function

0

$begingroup$

We have the following function:

$mu_n(p)=frac12(n-1)!p^n$, where $ngeq 3$ and $0 leq p leq 1$.

Now, we want to find lim$mu_n(fraccn)$ as n goes to infinity. where $c$ is a constant. Clearly, $c geq 0$ and $n geq c$. Moreover, we observe that $frac12(n-1)!$ goes to infinity as $n$ goes to infinity and $(fraccn)^n$ goes to $0$ as $n$ goes to infinity. Thus, we have an indeterminate form. I tried to use the L'Hopital's (after conversion) but it doesn't seem to work in this case. I have also put that function on a computer, to get a better idea where I might be going with this and it looks like for some (small) $c$'s the limit goes to $0$ and then for bigger $c$'s it goes to infinity, which I don't really understand. How can we deal with this analytically? In the question, there is also a hint that Stirling’s approximation may be useful. I'd really appreciate some help.

limits

share|cite|improve this question

edited Mar 19 at 11:56

Chris Godsil

11.7k21635

asked Mar 19 at 11:41

amator2357amator2357

939

$endgroup$

add a comment | 

0

$begingroup$

We have the following function:

$mu_n(p)=frac12(n-1)!p^n$, where $ngeq 3$ and $0 leq p leq 1$.

Now, we want to find lim$mu_n(fraccn)$ as n goes to infinity. where $c$ is a constant. Clearly, $c geq 0$ and $n geq c$. Moreover, we observe that $frac12(n-1)!$ goes to infinity as $n$ goes to infinity and $(fraccn)^n$ goes to $0$ as $n$ goes to infinity. Thus, we have an indeterminate form. I tried to use the L'Hopital's (after conversion) but it doesn't seem to work in this case. I have also put that function on a computer, to get a better idea where I might be going with this and it looks like for some (small) $c$'s the limit goes to $0$ and then for bigger $c$'s it goes to infinity, which I don't really understand. How can we deal with this analytically? In the question, there is also a hint that Stirling’s approximation may be useful. I'd really appreciate some help.

limits

share|cite|improve this question

edited Mar 19 at 11:56

Chris Godsil

11.7k21635

asked Mar 19 at 11:41

amator2357amator2357

939

$endgroup$

add a comment | 

$begingroup$

We have the following function:

$mu_n(p)=frac12(n-1)!p^n$, where $ngeq 3$ and $0 leq p leq 1$.

Now, we want to find lim$mu_n(fraccn)$ as n goes to infinity. where $c$ is a constant. Clearly, $c geq 0$ and $n geq c$. Moreover, we observe that $frac12(n-1)!$ goes to infinity as $n$ goes to infinity and $(fraccn)^n$ goes to $0$ as $n$ goes to infinity. Thus, we have an indeterminate form. I tried to use the L'Hopital's (after conversion) but it doesn't seem to work in this case. I have also put that function on a computer, to get a better idea where I might be going with this and it looks like for some (small) $c$'s the limit goes to $0$ and then for bigger $c$'s it goes to infinity, which I don't really understand. How can we deal with this analytically? In the question, there is also a hint that Stirling’s approximation may be useful. I'd really appreciate some help.

limits

share|cite|improve this question

edited Mar 19 at 11:56

Chris Godsil

11.7k21635

asked Mar 19 at 11:41

amator2357amator2357

939

$endgroup$

share|cite|improve this question

edited Mar 19 at 11:56

Chris Godsil

11.7k21635

asked Mar 19 at 11:41

amator2357amator2357

939

share|cite|improve this question

edited Mar 19 at 11:56

Chris Godsil

11.7k21635

asked Mar 19 at 11:41

amator2357amator2357

939

edited Mar 19 at 11:56

Chris Godsil

11.7k21635

edited Mar 19 at 11:56

Chris Godsil

11.7k21635

asked Mar 19 at 11:41

amator2357amator2357

939

asked Mar 19 at 11:41

amator2357amator2357

939

1 Answer
1

active

oldest

votes

3

$begingroup$

Hint: consider the series
$$sum(n−1)!fracc^nn^n$$
and apply the quotient test. If the series is convergent, $(n−1)!fracc^nn^nto 0$.

share|cite|improve this answer

answered Mar 19 at 11:50

Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla

34.9k42971

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ohh, ok. So we get that the limit of $|fraca_n+1a_n|$, where $a_n$ is the sequence inside the series above, is equal to $0$<$1$ so by the quotient test it is (absolutely) convergent and hence the sequence $a_n$ converges to $0$ as $n$ goes to infinity.
  $endgroup$
  – amator2357
  Mar 19 at 12:18
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you for your help.
  $endgroup$
  – amator2357
  Mar 19 at 12:24
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I just realized that I made a mistake. The limit of $|fraca_n+1a_n|$ is given by $fracce$. Is that correct?
  $endgroup$
  – amator2357
  Mar 19 at 12:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Assuming that it is correct and then using the quotient test we could say that $(n-1)!fracc^nn^n to 0$ when $c<e$. But how can we then deal with the cases when $c geq e$?
  $endgroup$
  – amator2357
  Mar 19 at 13:15
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @amator2357, if $c > e$ the general term $toinfty$.
  $endgroup$
  – Martín-Blas Pérez Pinilla
  Mar 19 at 14:05
  

  
  
  
  

 | 
show 5 more comments

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3

$begingroup$

Hint: consider the series
$$sum(n−1)!fracc^nn^n$$
and apply the quotient test. If the series is convergent, $(n−1)!fracc^nn^nto 0$.

share|cite|improve this answer

answered Mar 19 at 11:50

Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla

34.9k42971

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ohh, ok. So we get that the limit of $|fraca_n+1a_n|$, where $a_n$ is the sequence inside the series above, is equal to $0$<$1$ so by the quotient test it is (absolutely) convergent and hence the sequence $a_n$ converges to $0$ as $n$ goes to infinity.
  $endgroup$
  – amator2357
  Mar 19 at 12:18
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you for your help.
  $endgroup$
  – amator2357
  Mar 19 at 12:24
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I just realized that I made a mistake. The limit of $|fraca_n+1a_n|$ is given by $fracce$. Is that correct?
  $endgroup$
  – amator2357
  Mar 19 at 12:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Assuming that it is correct and then using the quotient test we could say that $(n-1)!fracc^nn^n to 0$ when $c<e$. But how can we then deal with the cases when $c geq e$?
  $endgroup$
  – amator2357
  Mar 19 at 13:15
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @amator2357, if $c > e$ the general term $toinfty$.
  $endgroup$
  – Martín-Blas Pérez Pinilla
  Mar 19 at 14:05
  

  
  
  
  

 | 
show 5 more comments

3

$begingroup$

Hint: consider the series
$$sum(n−1)!fracc^nn^n$$
and apply the quotient test. If the series is convergent, $(n−1)!fracc^nn^nto 0$.

share|cite|improve this answer

answered Mar 19 at 11:50

Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla

34.9k42971

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ohh, ok. So we get that the limit of $|fraca_n+1a_n|$, where $a_n$ is the sequence inside the series above, is equal to $0$<$1$ so by the quotient test it is (absolutely) convergent and hence the sequence $a_n$ converges to $0$ as $n$ goes to infinity.
  $endgroup$
  – amator2357
  Mar 19 at 12:18
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you for your help.
  $endgroup$
  – amator2357
  Mar 19 at 12:24
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I just realized that I made a mistake. The limit of $|fraca_n+1a_n|$ is given by $fracce$. Is that correct?
  $endgroup$
  – amator2357
  Mar 19 at 12:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Assuming that it is correct and then using the quotient test we could say that $(n-1)!fracc^nn^n to 0$ when $c<e$. But how can we then deal with the cases when $c geq e$?
  $endgroup$
  – amator2357
  Mar 19 at 13:15
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @amator2357, if $c > e$ the general term $toinfty$.
  $endgroup$
  – Martín-Blas Pérez Pinilla
  Mar 19 at 14:05
  

  
  
  
  

 | 
show 5 more comments

$begingroup$

Hint: consider the series
$$sum(n−1)!fracc^nn^n$$
and apply the quotient test. If the series is convergent, $(n−1)!fracc^nn^nto 0$.

share|cite|improve this answer

answered Mar 19 at 11:50

Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla

34.9k42971

$endgroup$

share|cite|improve this answer

answered Mar 19 at 11:50

Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla

34.9k42971

answered Mar 19 at 11:50

Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla

34.9k42971

answered Mar 19 at 11:50

Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla

34.9k42971