why is the limit of this expression equal to 1?












1












$begingroup$


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







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$endgroup$

















    1












    $begingroup$


    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







    share|cite|improve this question











    $endgroup$















      1












      1








      1


      2



      $begingroup$


      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







      share|cite|improve this question











      $endgroup$




      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






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 2 hours ago







      billyandr

















      asked 2 hours ago









      billyandrbillyandr

      155




      155






















          2 Answers
          2






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          5












          $begingroup$

          It is rather obvious if you cancel the factorials:



          $$frac{n!}{n^{k}(n-k)! } =frac{overbrace{n(n-1)cdots (n-k+1)}^{k; factors}}{n^k}= 1cdot left(1-frac{1}{n}right)cdots left(1-frac{k-1}{n}right)stackrel{n to infty}{longrightarrow} 1$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            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





















          2












          $begingroup$

          $$a_n=frac{n!}{n^{k}(n-k)! }implies log(a_n)=log(n!)-k log(n)-log((n-k)!)$$



          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
          $$a_n=e^{log(a_n)}=1+frac{k(1-k)}{2 n}+Oleft(frac{1}{n^2}right)$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            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














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          2 Answers
          2






          active

          oldest

          votes








          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          5












          $begingroup$

          It is rather obvious if you cancel the factorials:



          $$frac{n!}{n^{k}(n-k)! } =frac{overbrace{n(n-1)cdots (n-k+1)}^{k; factors}}{n^k}= 1cdot left(1-frac{1}{n}right)cdots left(1-frac{k-1}{n}right)stackrel{n to infty}{longrightarrow} 1$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            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


















          5












          $begingroup$

          It is rather obvious if you cancel the factorials:



          $$frac{n!}{n^{k}(n-k)! } =frac{overbrace{n(n-1)cdots (n-k+1)}^{k; factors}}{n^k}= 1cdot left(1-frac{1}{n}right)cdots left(1-frac{k-1}{n}right)stackrel{n to infty}{longrightarrow} 1$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            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
















          5












          5








          5





          $begingroup$

          It is rather obvious if you cancel the factorials:



          $$frac{n!}{n^{k}(n-k)! } =frac{overbrace{n(n-1)cdots (n-k+1)}^{k; factors}}{n^k}= 1cdot left(1-frac{1}{n}right)cdots left(1-frac{k-1}{n}right)stackrel{n to infty}{longrightarrow} 1$$






          share|cite|improve this answer









          $endgroup$



          It is rather obvious if you cancel the factorials:



          $$frac{n!}{n^{k}(n-k)! } =frac{overbrace{n(n-1)cdots (n-k+1)}^{k; factors}}{n^k}= 1cdot left(1-frac{1}{n}right)cdots left(1-frac{k-1}{n}right)stackrel{n to infty}{longrightarrow} 1$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          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




















          • $begingroup$
            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













          2












          $begingroup$

          $$a_n=frac{n!}{n^{k}(n-k)! }implies log(a_n)=log(n!)-k log(n)-log((n-k)!)$$



          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
          $$a_n=e^{log(a_n)}=1+frac{k(1-k)}{2 n}+Oleft(frac{1}{n^2}right)$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            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


















          2












          $begingroup$

          $$a_n=frac{n!}{n^{k}(n-k)! }implies log(a_n)=log(n!)-k log(n)-log((n-k)!)$$



          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
          $$a_n=e^{log(a_n)}=1+frac{k(1-k)}{2 n}+Oleft(frac{1}{n^2}right)$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            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
















          2












          2








          2





          $begingroup$

          $$a_n=frac{n!}{n^{k}(n-k)! }implies log(a_n)=log(n!)-k log(n)-log((n-k)!)$$



          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
          $$a_n=e^{log(a_n)}=1+frac{k(1-k)}{2 n}+Oleft(frac{1}{n^2}right)$$






          share|cite|improve this answer









          $endgroup$



          $$a_n=frac{n!}{n^{k}(n-k)! }implies log(a_n)=log(n!)-k log(n)-log((n-k)!)$$



          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
          $$a_n=e^{log(a_n)}=1+frac{k(1-k)}{2 n}+Oleft(frac{1}{n^2}right)$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          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




















          • $begingroup$
            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




















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