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why is the limit of this expression equal to 1?

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why is the limit of this expression equal to 1?

1

2

$begingroup$

I found something which I find confusing.

$$
lim_nrightarrow infty fracn!n^k(n-k)! =1
$$

It was something I encountered while learning probability on this webpage.

limits

share|cite|improve this question

edited 2 hours ago

billyandr

asked 2 hours ago

billyandrbillyandr

155

$endgroup$

add a comment | 

1

2

$begingroup$

I found something which I find confusing.

$$
lim_nrightarrow infty fracn!n^k(n-k)! =1
$$

It was something I encountered while learning probability on this webpage.

limits

share|cite|improve this question

edited 2 hours ago

billyandr

asked 2 hours ago

billyandrbillyandr

155

$endgroup$

add a comment | 

$begingroup$

I found something which I find confusing.

$$
lim_nrightarrow infty fracn!n^k(n-k)! =1
$$

It was something I encountered while learning probability on this webpage.

limits

share|cite|improve this question

edited 2 hours ago

billyandr

asked 2 hours ago

billyandrbillyandr

155

$endgroup$

share|cite|improve this question

edited 2 hours ago

billyandr

asked 2 hours ago

billyandrbillyandr

155

share|cite|improve this question

edited 2 hours ago

billyandr

asked 2 hours ago

billyandrbillyandr

155

asked 2 hours ago

billyandrbillyandr

155

asked 2 hours ago

billyandrbillyandr

155

2 Answers
2

active

oldest

votes

5

$begingroup$

It is rather obvious if you cancel the factorials:

$$fracn!n^k(n-k)! =fracoverbracen(n-1)cdots (n-k+1)^k; factorsn^k= 1cdot left(1-frac1nright)cdots left(1-frack-1nright)stackreln to inftylongrightarrow 1$$

share|cite|improve this answer

answered 1 hour ago

trancelocationtrancelocation

14.1k1829

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you so much. I didn't know it was right there under my eyes.
  $endgroup$
  – billyandr
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You are welcome. This "not seeing the obvious" just happens once in a while, I think, to all who do maths. So, it is good to have a math platform like this one. :-)
  $endgroup$
  – trancelocation
  1 hour ago
  
  

  
  
  
  

add a comment | 

2

$begingroup$

$$a_n=fracn!n^k(n-k)! implies log(a_n)=log(n!)-k log(n)-log((n-k)!)$$

Use Stirling approximation and continue with Taylor series to get
$$log(a_n)=frack(1-k)2 n+Oleft(frac1n^2right)$$ Continue with Taylor
$$a_n=e^log(a_n)=1+frack(1-k)2 n+Oleft(frac1n^2right)$$

share|cite|improve this answer

answered 1 hour ago

Claude LeiboviciClaude Leibovici

125k1158135

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This has already a slight touch of overkill, hasn't it? :-)
  $endgroup$
  – trancelocation
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @trancelocation. You are totally right for the limit. One of my manias is to always look at the approach to the limit. Have a look at matheducators.stackexchange.com/questions/8339/… . Cheers :-)
  $endgroup$
  – Claude Leibovici
  1 hour ago
  
  

  
  
  
  

add a comment | 

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5

$begingroup$

It is rather obvious if you cancel the factorials:

$$fracn!n^k(n-k)! =fracoverbracen(n-1)cdots (n-k+1)^k; factorsn^k= 1cdot left(1-frac1nright)cdots left(1-frack-1nright)stackreln to inftylongrightarrow 1$$

share|cite|improve this answer

answered 1 hour ago

trancelocationtrancelocation

14.1k1829

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you so much. I didn't know it was right there under my eyes.
  $endgroup$
  – billyandr
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You are welcome. This "not seeing the obvious" just happens once in a while, I think, to all who do maths. So, it is good to have a math platform like this one. :-)
  $endgroup$
  – trancelocation
  1 hour ago
  
  

  
  
  
  

add a comment | 

5

$begingroup$

It is rather obvious if you cancel the factorials:

$$fracn!n^k(n-k)! =fracoverbracen(n-1)cdots (n-k+1)^k; factorsn^k= 1cdot left(1-frac1nright)cdots left(1-frack-1nright)stackreln to inftylongrightarrow 1$$

share|cite|improve this answer

answered 1 hour ago

trancelocationtrancelocation

14.1k1829

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you so much. I didn't know it was right there under my eyes.
  $endgroup$
  – billyandr
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You are welcome. This "not seeing the obvious" just happens once in a while, I think, to all who do maths. So, it is good to have a math platform like this one. :-)
  $endgroup$
  – trancelocation
  1 hour ago
  
  

  
  
  
  

add a comment | 

$begingroup$

It is rather obvious if you cancel the factorials:

$$fracn!n^k(n-k)! =fracoverbracen(n-1)cdots (n-k+1)^k; factorsn^k= 1cdot left(1-frac1nright)cdots left(1-frack-1nright)stackreln to inftylongrightarrow 1$$

share|cite|improve this answer

answered 1 hour ago

trancelocationtrancelocation

14.1k1829

$endgroup$

share|cite|improve this answer

answered 1 hour ago

trancelocationtrancelocation

14.1k1829

answered 1 hour ago

trancelocationtrancelocation

14.1k1829

answered 1 hour ago

trancelocationtrancelocation

14.1k1829

2

$begingroup$

$$a_n=fracn!n^k(n-k)! implies log(a_n)=log(n!)-k log(n)-log((n-k)!)$$

Use Stirling approximation and continue with Taylor series to get
$$log(a_n)=frack(1-k)2 n+Oleft(frac1n^2right)$$ Continue with Taylor
$$a_n=e^log(a_n)=1+frack(1-k)2 n+Oleft(frac1n^2right)$$

share|cite|improve this answer

answered 1 hour ago

Claude LeiboviciClaude Leibovici

125k1158135

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This has already a slight touch of overkill, hasn't it? :-)
  $endgroup$
  – trancelocation
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @trancelocation. You are totally right for the limit. One of my manias is to always look at the approach to the limit. Have a look at matheducators.stackexchange.com/questions/8339/… . Cheers :-)
  $endgroup$
  – Claude Leibovici
  1 hour ago
  
  

  
  
  
  

add a comment | 

2

$begingroup$

$$a_n=fracn!n^k(n-k)! implies log(a_n)=log(n!)-k log(n)-log((n-k)!)$$

Use Stirling approximation and continue with Taylor series to get
$$log(a_n)=frack(1-k)2 n+Oleft(frac1n^2right)$$ Continue with Taylor
$$a_n=e^log(a_n)=1+frack(1-k)2 n+Oleft(frac1n^2right)$$

share|cite|improve this answer

answered 1 hour ago

Claude LeiboviciClaude Leibovici

125k1158135

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This has already a slight touch of overkill, hasn't it? :-)
  $endgroup$
  – trancelocation
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @trancelocation. You are totally right for the limit. One of my manias is to always look at the approach to the limit. Have a look at matheducators.stackexchange.com/questions/8339/… . Cheers :-)
  $endgroup$
  – Claude Leibovici
  1 hour ago
  
  

  
  
  
  

add a comment | 

$begingroup$

$$a_n=fracn!n^k(n-k)! implies log(a_n)=log(n!)-k log(n)-log((n-k)!)$$

Use Stirling approximation and continue with Taylor series to get
$$log(a_n)=frack(1-k)2 n+Oleft(frac1n^2right)$$ Continue with Taylor
$$a_n=e^log(a_n)=1+frack(1-k)2 n+Oleft(frac1n^2right)$$

share|cite|improve this answer

answered 1 hour ago

Claude LeiboviciClaude Leibovici

125k1158135

$endgroup$

share|cite|improve this answer

answered 1 hour ago

Claude LeiboviciClaude Leibovici

125k1158135

answered 1 hour ago

Claude LeiboviciClaude Leibovici

125k1158135

answered 1 hour ago

Claude LeiboviciClaude Leibovici

125k1158135