Limit of $frac{(n-1)^{2n-1-k}}{n^n (n-1-k)^{n-1-k}}$












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I'm trying to calculate the limit of $$frac{(n-1)^{2n-1-k}}{n^n (n-1-k)^{n-1-k}}$$ as $n to infty$. I know (using Wolfram Alpha) that the limit should be equal to $e^{k-1}$. However I'm unable to manipulate the expression in the way that I could see the limit to be true.






limits







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    I'm trying to calculate the limit of $$frac{(n-1)^{2n-1-k}}{n^n (n-1-k)^{n-1-k}}$$ as $n to infty$. I know (using Wolfram Alpha) that the limit should be equal to $e^{k-1}$. However I'm unable to manipulate the expression in the way that I could see the limit to be true.






    limits







    share|cite|improve this question



























      2












      2








      2







      I'm trying to calculate the limit of $$frac{(n-1)^{2n-1-k}}{n^n (n-1-k)^{n-1-k}}$$ as $n to infty$. I know (using Wolfram Alpha) that the limit should be equal to $e^{k-1}$. However I'm unable to manipulate the expression in the way that I could see the limit to be true.






      limits







      share|cite|improve this question















      I'm trying to calculate the limit of $$frac{(n-1)^{2n-1-k}}{n^n (n-1-k)^{n-1-k}}$$ as $n to infty$. I know (using Wolfram Alpha) that the limit should be equal to $e^{k-1}$. However I'm unable to manipulate the expression in the way that I could see the limit to be true.






      limits




      limits






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      edited Nov 27 at 12:23









      Martin Sleziak

      44.7k7115270




      44.7k7115270










      asked Oct 30 at 12:47









      Kplusn

      1199




      1199






















          2 Answers
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          Just split it up appropriately. A possible way is as follows:



          $$frac{(n-1)^{2n-1-k}}{n^n (n-1-k)^{n-1-k}} = frac{(n-1)^{2n}cdot (n-1-k)^{k+1}}{n^ncdot (n-1-k)^n cdot (n-1)^{k+1}} $$ $$ = underbrace{frac{(n-1-k)^{k+1}}{(n-1)^{k+1}}}_{stackrel{n to infty}{longrightarrow}1}cdot underbrace{frac{(n-1)^n}{n^n}}_{stackrel{n to infty}{longrightarrow}e^{-1}}cdot underbrace{frac{1}{frac{(n-1-k)^n}{(n-1)^n}}}_{stackrel{n to infty}{longrightarrow}frac{1}{e^{-k}} =e^{k}}$$






          share|cite|improve this answer





















          • Great! I tried to do it in a similar way but seems that Im not that good of a splitter.
            – Kplusn
            Oct 30 at 13:08



















          1














          HINT



          We have



          $$dfrac{(n-1)^{2n-1-k}}{n^n (n-1-k)^{n-1-k}}=dfrac{(n-1)^n}{n^n }dfrac{(n-1)^{n-1-k}}{ (n-1-k)^{n-1-k}}=left( 1-frac1n right)^nfrac{1}{left( 1-frac k{n-1} right)^{n-1-k}}$$






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            2 Answers
            2






            active

            oldest

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            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            4














            Just split it up appropriately. A possible way is as follows:



            $$frac{(n-1)^{2n-1-k}}{n^n (n-1-k)^{n-1-k}} = frac{(n-1)^{2n}cdot (n-1-k)^{k+1}}{n^ncdot (n-1-k)^n cdot (n-1)^{k+1}} $$ $$ = underbrace{frac{(n-1-k)^{k+1}}{(n-1)^{k+1}}}_{stackrel{n to infty}{longrightarrow}1}cdot underbrace{frac{(n-1)^n}{n^n}}_{stackrel{n to infty}{longrightarrow}e^{-1}}cdot underbrace{frac{1}{frac{(n-1-k)^n}{(n-1)^n}}}_{stackrel{n to infty}{longrightarrow}frac{1}{e^{-k}} =e^{k}}$$






            share|cite|improve this answer





















            • Great! I tried to do it in a similar way but seems that Im not that good of a splitter.
              – Kplusn
              Oct 30 at 13:08
















            4














            Just split it up appropriately. A possible way is as follows:



            $$frac{(n-1)^{2n-1-k}}{n^n (n-1-k)^{n-1-k}} = frac{(n-1)^{2n}cdot (n-1-k)^{k+1}}{n^ncdot (n-1-k)^n cdot (n-1)^{k+1}} $$ $$ = underbrace{frac{(n-1-k)^{k+1}}{(n-1)^{k+1}}}_{stackrel{n to infty}{longrightarrow}1}cdot underbrace{frac{(n-1)^n}{n^n}}_{stackrel{n to infty}{longrightarrow}e^{-1}}cdot underbrace{frac{1}{frac{(n-1-k)^n}{(n-1)^n}}}_{stackrel{n to infty}{longrightarrow}frac{1}{e^{-k}} =e^{k}}$$






            share|cite|improve this answer





















            • Great! I tried to do it in a similar way but seems that Im not that good of a splitter.
              – Kplusn
              Oct 30 at 13:08














            4












            4








            4






            Just split it up appropriately. A possible way is as follows:



            $$frac{(n-1)^{2n-1-k}}{n^n (n-1-k)^{n-1-k}} = frac{(n-1)^{2n}cdot (n-1-k)^{k+1}}{n^ncdot (n-1-k)^n cdot (n-1)^{k+1}} $$ $$ = underbrace{frac{(n-1-k)^{k+1}}{(n-1)^{k+1}}}_{stackrel{n to infty}{longrightarrow}1}cdot underbrace{frac{(n-1)^n}{n^n}}_{stackrel{n to infty}{longrightarrow}e^{-1}}cdot underbrace{frac{1}{frac{(n-1-k)^n}{(n-1)^n}}}_{stackrel{n to infty}{longrightarrow}frac{1}{e^{-k}} =e^{k}}$$






            share|cite|improve this answer












            Just split it up appropriately. A possible way is as follows:



            $$frac{(n-1)^{2n-1-k}}{n^n (n-1-k)^{n-1-k}} = frac{(n-1)^{2n}cdot (n-1-k)^{k+1}}{n^ncdot (n-1-k)^n cdot (n-1)^{k+1}} $$ $$ = underbrace{frac{(n-1-k)^{k+1}}{(n-1)^{k+1}}}_{stackrel{n to infty}{longrightarrow}1}cdot underbrace{frac{(n-1)^n}{n^n}}_{stackrel{n to infty}{longrightarrow}e^{-1}}cdot underbrace{frac{1}{frac{(n-1-k)^n}{(n-1)^n}}}_{stackrel{n to infty}{longrightarrow}frac{1}{e^{-k}} =e^{k}}$$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Oct 30 at 13:01









            trancelocation

            9,1051521




            9,1051521












            • Great! I tried to do it in a similar way but seems that Im not that good of a splitter.
              – Kplusn
              Oct 30 at 13:08


















            • Great! I tried to do it in a similar way but seems that Im not that good of a splitter.
              – Kplusn
              Oct 30 at 13:08
















            Great! I tried to do it in a similar way but seems that Im not that good of a splitter.
            – Kplusn
            Oct 30 at 13:08




            Great! I tried to do it in a similar way but seems that Im not that good of a splitter.
            – Kplusn
            Oct 30 at 13:08











            1














            HINT



            We have



            $$dfrac{(n-1)^{2n-1-k}}{n^n (n-1-k)^{n-1-k}}=dfrac{(n-1)^n}{n^n }dfrac{(n-1)^{n-1-k}}{ (n-1-k)^{n-1-k}}=left( 1-frac1n right)^nfrac{1}{left( 1-frac k{n-1} right)^{n-1-k}}$$






            share|cite|improve this answer




























              1














              HINT



              We have



              $$dfrac{(n-1)^{2n-1-k}}{n^n (n-1-k)^{n-1-k}}=dfrac{(n-1)^n}{n^n }dfrac{(n-1)^{n-1-k}}{ (n-1-k)^{n-1-k}}=left( 1-frac1n right)^nfrac{1}{left( 1-frac k{n-1} right)^{n-1-k}}$$






              share|cite|improve this answer


























                1












                1








                1






                HINT



                We have



                $$dfrac{(n-1)^{2n-1-k}}{n^n (n-1-k)^{n-1-k}}=dfrac{(n-1)^n}{n^n }dfrac{(n-1)^{n-1-k}}{ (n-1-k)^{n-1-k}}=left( 1-frac1n right)^nfrac{1}{left( 1-frac k{n-1} right)^{n-1-k}}$$






                share|cite|improve this answer














                HINT



                We have



                $$dfrac{(n-1)^{2n-1-k}}{n^n (n-1-k)^{n-1-k}}=dfrac{(n-1)^n}{n^n }dfrac{(n-1)^{n-1-k}}{ (n-1-k)^{n-1-k}}=left( 1-frac1n right)^nfrac{1}{left( 1-frac k{n-1} right)^{n-1-k}}$$







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Oct 30 at 13:09

























                answered Oct 30 at 13:02









                gimusi

                1




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