Closed form for a summation.












1














Mathematica says $displaystylesum_{k=1}^{n} dfrac{binom{n-1}{k-1}}{n^k} = dfrac{(n+1)^{n-1}}{n^n}$. I would like to know how to show this equality.



The left hand side is the solution to a probability problem: Select integers (with replacement) from {1,2,...,n} until the cumulative sum of your selected integers is at least n. What is the probability that your cumulative sum is exactly n? The right hand side is the ratio of labeled trees on n+1 nodes to functions from [n] into [n]. A combinatorial proof to the equations would be great. I would settle for an analytic proof.






probability combinatorics







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  • 1




    You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an edit): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance.
    – Shaun
    Nov 26 at 16:16






  • 2




    Expand $(n+1)^{n-1}$.
    – Yves Daoust
    Nov 26 at 16:21










  • @ Yves Daoust. Yes Thank you. I see it now.
    – Geoffrey Critzer
    Nov 26 at 16:33
















1














Mathematica says $displaystylesum_{k=1}^{n} dfrac{binom{n-1}{k-1}}{n^k} = dfrac{(n+1)^{n-1}}{n^n}$. I would like to know how to show this equality.



The left hand side is the solution to a probability problem: Select integers (with replacement) from {1,2,...,n} until the cumulative sum of your selected integers is at least n. What is the probability that your cumulative sum is exactly n? The right hand side is the ratio of labeled trees on n+1 nodes to functions from [n] into [n]. A combinatorial proof to the equations would be great. I would settle for an analytic proof.






probability combinatorics







share|cite|improve this question




















  • 1




    You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an edit): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance.
    – Shaun
    Nov 26 at 16:16






  • 2




    Expand $(n+1)^{n-1}$.
    – Yves Daoust
    Nov 26 at 16:21










  • @ Yves Daoust. Yes Thank you. I see it now.
    – Geoffrey Critzer
    Nov 26 at 16:33














1












1








1


1





Mathematica says $displaystylesum_{k=1}^{n} dfrac{binom{n-1}{k-1}}{n^k} = dfrac{(n+1)^{n-1}}{n^n}$. I would like to know how to show this equality.



The left hand side is the solution to a probability problem: Select integers (with replacement) from {1,2,...,n} until the cumulative sum of your selected integers is at least n. What is the probability that your cumulative sum is exactly n? The right hand side is the ratio of labeled trees on n+1 nodes to functions from [n] into [n]. A combinatorial proof to the equations would be great. I would settle for an analytic proof.






probability combinatorics







share|cite|improve this question















Mathematica says $displaystylesum_{k=1}^{n} dfrac{binom{n-1}{k-1}}{n^k} = dfrac{(n+1)^{n-1}}{n^n}$. I would like to know how to show this equality.



The left hand side is the solution to a probability problem: Select integers (with replacement) from {1,2,...,n} until the cumulative sum of your selected integers is at least n. What is the probability that your cumulative sum is exactly n? The right hand side is the ratio of labeled trees on n+1 nodes to functions from [n] into [n]. A combinatorial proof to the equations would be great. I would settle for an analytic proof.






probability combinatorics




probability combinatorics






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 26 at 16:31









Mike Earnest

20.1k11950




20.1k11950










asked Nov 26 at 16:11









Geoffrey Critzer

1,5001127




1,5001127








  • 1




    You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an edit): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance.
    – Shaun
    Nov 26 at 16:16






  • 2




    Expand $(n+1)^{n-1}$.
    – Yves Daoust
    Nov 26 at 16:21










  • @ Yves Daoust. Yes Thank you. I see it now.
    – Geoffrey Critzer
    Nov 26 at 16:33














  • 1




    You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an edit): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance.
    – Shaun
    Nov 26 at 16:16






  • 2




    Expand $(n+1)^{n-1}$.
    – Yves Daoust
    Nov 26 at 16:21










  • @ Yves Daoust. Yes Thank you. I see it now.
    – Geoffrey Critzer
    Nov 26 at 16:33








1




1




You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an edit): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance.
– Shaun
Nov 26 at 16:16




You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an edit): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance.
– Shaun
Nov 26 at 16:16




2




2




Expand $(n+1)^{n-1}$.
– Yves Daoust
Nov 26 at 16:21




Expand $(n+1)^{n-1}$.
– Yves Daoust
Nov 26 at 16:21












@ Yves Daoust. Yes Thank you. I see it now.
– Geoffrey Critzer
Nov 26 at 16:33




@ Yves Daoust. Yes Thank you. I see it now.
– Geoffrey Critzer
Nov 26 at 16:33










1 Answer
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According to the Binomial Theorem we get as a series expansion for $(n+1)^{n-1}$ the following



$$(1+n)^{n-1}=sum_{k=0}^{n-1}binom{n-1}{k}n^{n-1-k}$$



By applying the index shift $k'=k+1$ - and therefore $k=k'-1$ - we further get



$$sum_{k=0}^{n-1}binom{n-1}{k}n^{n-1-k}=sum_{k'=1}^{n}binom{n-1}{k'-1}n^{n-k'}$$



One may note that within the term $n^{n-k'}$ the $n^n$ is independed of the index. By reshaping the whole equality we get



$$begin{align}
(1+n)^{n-1}&=sum_{k'=1}^{n}binom{n-1}{k'-1}n^{n-k'}\
&=n^nsum_{k'=1}^{n}frac{binom{n-1}{k'-1}}{n^{k'}}\
Leftrightarrow frac{(1+n)^{n-1}}{n^n}&=sum_{k'=1}^{n}frac{binom{n-1}{k'-1}}{n^{k'}}
end{align}$$



which is your desired result.






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    According to the Binomial Theorem we get as a series expansion for $(n+1)^{n-1}$ the following



    $$(1+n)^{n-1}=sum_{k=0}^{n-1}binom{n-1}{k}n^{n-1-k}$$



    By applying the index shift $k'=k+1$ - and therefore $k=k'-1$ - we further get



    $$sum_{k=0}^{n-1}binom{n-1}{k}n^{n-1-k}=sum_{k'=1}^{n}binom{n-1}{k'-1}n^{n-k'}$$



    One may note that within the term $n^{n-k'}$ the $n^n$ is independed of the index. By reshaping the whole equality we get



    $$begin{align}
    (1+n)^{n-1}&=sum_{k'=1}^{n}binom{n-1}{k'-1}n^{n-k'}\
    &=n^nsum_{k'=1}^{n}frac{binom{n-1}{k'-1}}{n^{k'}}\
    Leftrightarrow frac{(1+n)^{n-1}}{n^n}&=sum_{k'=1}^{n}frac{binom{n-1}{k'-1}}{n^{k'}}
    end{align}$$



    which is your desired result.






    share|cite|improve this answer




























      2














      According to the Binomial Theorem we get as a series expansion for $(n+1)^{n-1}$ the following



      $$(1+n)^{n-1}=sum_{k=0}^{n-1}binom{n-1}{k}n^{n-1-k}$$



      By applying the index shift $k'=k+1$ - and therefore $k=k'-1$ - we further get



      $$sum_{k=0}^{n-1}binom{n-1}{k}n^{n-1-k}=sum_{k'=1}^{n}binom{n-1}{k'-1}n^{n-k'}$$



      One may note that within the term $n^{n-k'}$ the $n^n$ is independed of the index. By reshaping the whole equality we get



      $$begin{align}
      (1+n)^{n-1}&=sum_{k'=1}^{n}binom{n-1}{k'-1}n^{n-k'}\
      &=n^nsum_{k'=1}^{n}frac{binom{n-1}{k'-1}}{n^{k'}}\
      Leftrightarrow frac{(1+n)^{n-1}}{n^n}&=sum_{k'=1}^{n}frac{binom{n-1}{k'-1}}{n^{k'}}
      end{align}$$



      which is your desired result.






      share|cite|improve this answer


























        2












        2








        2






        According to the Binomial Theorem we get as a series expansion for $(n+1)^{n-1}$ the following



        $$(1+n)^{n-1}=sum_{k=0}^{n-1}binom{n-1}{k}n^{n-1-k}$$



        By applying the index shift $k'=k+1$ - and therefore $k=k'-1$ - we further get



        $$sum_{k=0}^{n-1}binom{n-1}{k}n^{n-1-k}=sum_{k'=1}^{n}binom{n-1}{k'-1}n^{n-k'}$$



        One may note that within the term $n^{n-k'}$ the $n^n$ is independed of the index. By reshaping the whole equality we get



        $$begin{align}
        (1+n)^{n-1}&=sum_{k'=1}^{n}binom{n-1}{k'-1}n^{n-k'}\
        &=n^nsum_{k'=1}^{n}frac{binom{n-1}{k'-1}}{n^{k'}}\
        Leftrightarrow frac{(1+n)^{n-1}}{n^n}&=sum_{k'=1}^{n}frac{binom{n-1}{k'-1}}{n^{k'}}
        end{align}$$



        which is your desired result.






        share|cite|improve this answer














        According to the Binomial Theorem we get as a series expansion for $(n+1)^{n-1}$ the following



        $$(1+n)^{n-1}=sum_{k=0}^{n-1}binom{n-1}{k}n^{n-1-k}$$



        By applying the index shift $k'=k+1$ - and therefore $k=k'-1$ - we further get



        $$sum_{k=0}^{n-1}binom{n-1}{k}n^{n-1-k}=sum_{k'=1}^{n}binom{n-1}{k'-1}n^{n-k'}$$



        One may note that within the term $n^{n-k'}$ the $n^n$ is independed of the index. By reshaping the whole equality we get



        $$begin{align}
        (1+n)^{n-1}&=sum_{k'=1}^{n}binom{n-1}{k'-1}n^{n-k'}\
        &=n^nsum_{k'=1}^{n}frac{binom{n-1}{k'-1}}{n^{k'}}\
        Leftrightarrow frac{(1+n)^{n-1}}{n^n}&=sum_{k'=1}^{n}frac{binom{n-1}{k'-1}}{n^{k'}}
        end{align}$$



        which is your desired result.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Nov 26 at 17:17

























        answered Nov 26 at 16:34









        mrtaurho

        3,3982932




        3,3982932






























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