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why is the limit of this expression equal to 1?
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Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Finding the limit of the following expressionReforming series expression for limit of e$lim_{x rightarrow infty}left(frac{pi}{2}-tan^{-1}xright)^{Largefrac{1}{x}}$ Why aren't these two limits equal when they should be?What is the value of this limit?limit of an expressionUsing a definite integral find the value of $lim_{nrightarrow infty }(frac{1}{n}+frac{1}{n+1}+…+frac{1}{2n})$Why is the following limit operation valid?Is this expression on limit valid and/or meaningful?Why does this limit equal 0?A Problem on the Limit of an Integral
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I found something which I find confusing.
$$
lim_{nrightarrow infty} frac{n!}{n^{k}(n-k)! } =1
$$
It was something I encountered while learning probability on this webpage.
limits
asked 2 hours ago
billyandrbillyandr
155
$endgroup$
add a comment |
$begingroup$
I found something which I find confusing.
$$
lim_{nrightarrow infty} frac{n!}{n^{k}(n-k)! } =1
$$
It was something I encountered while learning probability on this webpage.
limits
asked 2 hours ago
billyandrbillyandr
155
$endgroup$
add a comment |
$begingroup$
I found something which I find confusing.
$$
lim_{nrightarrow infty} frac{n!}{n^{k}(n-k)! } =1
$$
It was something I encountered while learning probability on this webpage.
limits
asked 2 hours ago
billyandrbillyandr
155
$endgroup$
I found something which I find confusing.
$$
lim_{nrightarrow infty} frac{n!}{n^{k}(n-k)! } =1
$$
It was something I encountered while learning probability on this webpage.
limits
limits
asked 2 hours ago
billyandrbillyandr
155
asked 2 hours ago
billyandrbillyandr
155
edited 2 hours ago
billyandr
asked 2 hours ago
billyandrbillyandr
155
asked 2 hours ago
billyandrbillyandr
155
asked 2 hours ago
billyandrbillyandr
155
155
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
It is rather obvious if you cancel the factorials:
$$frac{n!}{n^{k}(n-k)! } =frac{overbrace{n(n-1)cdots (n-k+1)}^{k; factors}}{n^k}= 1cdot left(1-frac{1}{n}right)cdots left(1-frac{k-1}{n}right)stackrel{n to infty}{longrightarrow} 1$$
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 |
$begingroup$
$$a_n=frac{n!}{n^{k}(n-k)! }implies log(a_n)=log(n!)-k log(n)-log((n-k)!)$$
Use Stirling approximation and continue with Taylor series to get
$$log(a_n)=frac{k(1-k)}{2 n}+Oleft(frac{1}{n^2}right)$$ Continue with Taylor
$$a_n=e^{log(a_n)}=1+frac{k(1-k)}{2 n}+Oleft(frac{1}{n^2}right)$$
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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2 Answers
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active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
It is rather obvious if you cancel the factorials:
$$frac{n!}{n^{k}(n-k)! } =frac{overbrace{n(n-1)cdots (n-k+1)}^{k; factors}}{n^k}= 1cdot left(1-frac{1}{n}right)cdots left(1-frac{k-1}{n}right)stackrel{n to infty}{longrightarrow} 1$$
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:
$$frac{n!}{n^{k}(n-k)! } =frac{overbrace{n(n-1)cdots (n-k+1)}^{k; factors}}{n^k}= 1cdot left(1-frac{1}{n}right)cdots left(1-frac{k-1}{n}right)stackrel{n to infty}{longrightarrow} 1$$
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:
$$frac{n!}{n^{k}(n-k)! } =frac{overbrace{n(n-1)cdots (n-k+1)}^{k; factors}}{n^k}= 1cdot left(1-frac{1}{n}right)cdots left(1-frac{k-1}{n}right)stackrel{n to infty}{longrightarrow} 1$$
answered 1 hour ago
trancelocationtrancelocation
14.1k1829
$endgroup$
It is rather obvious if you cancel the factorials:
$$frac{n!}{n^{k}(n-k)! } =frac{overbrace{n(n-1)cdots (n-k+1)}^{k; factors}}{n^k}= 1cdot left(1-frac{1}{n}right)cdots left(1-frac{k-1}{n}right)stackrel{n to infty}{longrightarrow} 1$$
answered 1 hour ago
trancelocationtrancelocation
14.1k1829
answered 1 hour ago
trancelocationtrancelocation
14.1k1829
answered 1 hour ago
trancelocationtrancelocation
14.1k1829
answered 1 hour ago
trancelocationtrancelocation
14.1k1829
14.1k1829
$begingroup$
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 |
$begingroup$
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
$begingroup$
Thank you so much. I didn't know it was right there under my eyes.
$endgroup$
– billyandr
1 hour ago
$begingroup$
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
$begingroup$
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 |
$begingroup$
$$a_n=frac{n!}{n^{k}(n-k)! }implies log(a_n)=log(n!)-k log(n)-log((n-k)!)$$
Use Stirling approximation and continue with Taylor series to get
$$log(a_n)=frac{k(1-k)}{2 n}+Oleft(frac{1}{n^2}right)$$ Continue with Taylor
$$a_n=e^{log(a_n)}=1+frac{k(1-k)}{2 n}+Oleft(frac{1}{n^2}right)$$
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 |
$begingroup$
$$a_n=frac{n!}{n^{k}(n-k)! }implies log(a_n)=log(n!)-k log(n)-log((n-k)!)$$
Use Stirling approximation and continue with Taylor series to get
$$log(a_n)=frac{k(1-k)}{2 n}+Oleft(frac{1}{n^2}right)$$ Continue with Taylor
$$a_n=e^{log(a_n)}=1+frac{k(1-k)}{2 n}+Oleft(frac{1}{n^2}right)$$
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 |
$begingroup$
$$a_n=frac{n!}{n^{k}(n-k)! }implies log(a_n)=log(n!)-k log(n)-log((n-k)!)$$
Use Stirling approximation and continue with Taylor series to get
$$log(a_n)=frac{k(1-k)}{2 n}+Oleft(frac{1}{n^2}right)$$ Continue with Taylor
$$a_n=e^{log(a_n)}=1+frac{k(1-k)}{2 n}+Oleft(frac{1}{n^2}right)$$
answered 1 hour ago
Claude LeiboviciClaude Leibovici
125k1158135
$endgroup$
$$a_n=frac{n!}{n^{k}(n-k)! }implies log(a_n)=log(n!)-k log(n)-log((n-k)!)$$
Use Stirling approximation and continue with Taylor series to get
$$log(a_n)=frac{k(1-k)}{2 n}+Oleft(frac{1}{n^2}right)$$ Continue with Taylor
$$a_n=e^{log(a_n)}=1+frac{k(1-k)}{2 n}+Oleft(frac{1}{n^2}right)$$
answered 1 hour ago
Claude LeiboviciClaude Leibovici
125k1158135
answered 1 hour ago
Claude LeiboviciClaude Leibovici
125k1158135
answered 1 hour ago
Claude LeiboviciClaude Leibovici
125k1158135
answered 1 hour ago
Claude LeiboviciClaude Leibovici
125k1158135
125k1158135
$begingroup$
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 |
$begingroup$
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
$begingroup$
This has already a slight touch of overkill, hasn't it? :-)
$endgroup$
– trancelocation
1 hour ago
$begingroup$
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
$begingroup$
@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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