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Thread: Factorial gamma function

  1. #1
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    Factorial gamma function

    This is for high school but I suppose it is not studied in most high schools so I'll post it here. I don't know where to begin.

    Prove that $\displaystyle \sum _{n=1}^N \frac{(n+m)!}{n!}=-\frac{\Gamma (1+m) \Gamma (1+N)+m \Gamma (1+m) \Gamma (1+N)-\Gamma (2+m+N)}{(1+m) \Gamma (1+N)}$

    Any help will be appreciated
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by fobos3 View Post
    This is for high school but I suppose it is not studied in most high schools so I'll post it here. I don't know where to begin.

    Prove that $\displaystyle \sum _{n=1}^N \frac{(n+m)!}{n!}=-\frac{\Gamma (1+m) \Gamma (1+N)+m \Gamma (1+m) \Gamma (1+N)-\Gamma (2+m+N)}{(1+m) \Gamma (1+N)}$

    Any help will be appreciated
    From looking at it, I think the best way would be to prove this by mathematical induction:

    First, show that this is true for N=1:

    $\displaystyle \frac{(1+m)!}{1!}=-\frac{m! \Gamma (2)+m\cdot m! \Gamma (2)-(m+2)!}{(1+m) \Gamma (2)}$

    But $\displaystyle \Gamma(2)=1$, Thus $\displaystyle (1+m)!=-\frac{m! +(m+1)! -(m+2)!}{(1+m)}$

    Now, $\displaystyle -\frac{m! +(m+1)! -(m+2)!}{(1+m)}=\frac{-1-m-1+m^2+3m+2}{m+1}m!=\frac{m^2+2m+1}{m+1}m!$ $\displaystyle =\frac{(m+1)^2}{m+1}m!=(m+1)m!=(m+1)!$

    ------------------------------------

    Since we've shown this holds for N=1, Now we assume it holds for N=K

    The hard [and really messy part] is to show it holds for N=k+1.

    This means that $\displaystyle \sum _{n=1}^{k+1} \frac{(n+m)!}{n!}=-\frac{\Gamma (1+m) \Gamma (k+2)+m \Gamma (1+m) \Gamma (k+2)-\Gamma (k+3+m)}{(1+m) \Gamma (k+2)}$

    Note that $\displaystyle \sum_{n=1}^{k+1}\frac{(n+m)!}{n!}=\sum_{n=1}^k\fra c{(n+m)!}{n!}+\frac{(k+1+m)!}{(k+1)!}$

    So $\displaystyle \sum_{n=1}^k\frac{(n+m)!}{n!}$ $\displaystyle +\frac{(k+1+m)!}{(k+1)!}=-\frac{\Gamma (1+m) \Gamma (k+2)+m \Gamma (1+m) \Gamma (k+2)-\Gamma (k+3+m)}{(1+m) \Gamma (k+2)}$

    Knowing that $\displaystyle \Gamma(u)=(u-1)!$, we see that $\displaystyle \sum_{n=1}^k\frac{(n+m)!}{n!}$ $\displaystyle +\frac{(k+1+m)!}{(k+1)!}=-\frac{m!(k+1)!+m!m(k+1)!-(k+2+m)!}{(1+m)(k+1)!}$

    We can further simplify $\displaystyle -\frac{m!(k+1)!+m!m(k+1)!-(k+2+m)!}{(1+m)(k+1)!}=-\frac{(1+m)m!(k+1)!-(k+2+m)!}{(1+m)(k+1)!}$ $\displaystyle =-\frac{(m+1)!(k+1)!-(k+2+m)!}{(1+m)(k+1)!}$

    Since $\displaystyle \sum _{n=1}^k \frac{(n+m)!}{n!}=-\frac{\Gamma (1+m) \Gamma (1+k)+m \Gamma (1+m) \Gamma (1+k)-\Gamma (2+m+k)}{(1+m) \Gamma (1+k)}$ $\displaystyle =-\frac{m!k!+m!k!m-(k+1+m)!}{(1+m) k!}$

    Thus, we need to show that $\displaystyle -\frac{m!k!+m!k!m-(k+1+m)!}{(1+m) k!}+\frac{(k+1+m)!}{(k+1)!}$ $\displaystyle =\color{red}-\frac{(m+1)!(k+1)!-(k+2+m)!}{(1+m)(k+1)!}$

    Let's get the left side to look like the right side:

    $\displaystyle -\frac{m!k!+m!k!m-(k+1+m)!}{(1+m) k!}+\frac{(k+1+m)!}{(k+1)!}$ $\displaystyle =-\frac{(1+m)m!k!-(k+1+m)!}{(1+m)(1+k) k!}(k+1)+\frac{(1+m)(k+1+m)!}{(1+m)(k+1)k!}$

    $\displaystyle =\frac{-(m+1)!(k+1)!+(k+1)(k+1+m)!+(m+1)(k+1+m)!}{(m+1)(k+ 1)!}$ $\displaystyle =\frac{-(m+1)!(k+1)!+(k+2+m)(k+1+m)!}{(m+1)(k+1)!}$

    $\displaystyle =\frac{-(m+1)!(k+1)!+(k+2+m)!}{(m+1)(k+1)!}=\color{red}-\frac{(m+1)!(k+1)!-(k+2+m)!}{(m+1)(k+1)!}$

    This completes the inductive step.



    Does this make sense? Hopefully you can follow what I did...

    --Chris

    P.S. I wonder if anyone knows of a shorter way!?!?!?!
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  3. #3
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    Quote Originally Posted by Chris L T521 View Post
    From looking at it, I think the best way would be to prove this by mathematical induction:

    First, show that this is true for N=1:

    $\displaystyle \frac{(1+m)!}{1!}=-\frac{m! \Gamma (2)+m\cdot m! \Gamma (2)-(m+2)!}{(1+m) \Gamma (2)}$

    But $\displaystyle \Gamma(2)=1$, Thus $\displaystyle (1+m)!=-\frac{m! +(m+1)! -(m+2)!}{(1+m)}$

    Now, $\displaystyle -\frac{m! +(m+1)! -(m+2)!}{(1+m)}=\frac{-1-m-1+m^2+3m+2}{m+1}m!=\frac{m^2+2m+1}{m+1}m!$ $\displaystyle =\frac{(m+1)^2}{m+1}m!=(m+1)m!=(m+1)!$

    ------------------------------------

    Since we've shown this holds for N=1, Now we assume it holds for N=K

    The hard [and really messy part] is to show it holds for N=k+1.

    This means that $\displaystyle \sum _{n=1}^{k+1} \frac{(n+m)!}{n!}=-\frac{\Gamma (1+m) \Gamma (k+2)+m \Gamma (1+m) \Gamma (k+2)-\Gamma (k+3+m)}{(1+m) \Gamma (k+2)}$

    Note that $\displaystyle \sum_{n=1}^{k+1}\frac{(n+m)!}{n!}=\sum_{n=1}^k\fra c{(n+m)!}{n!}+\frac{(k+1+m)!}{(k+1)!}$

    So $\displaystyle \sum_{n=1}^k\frac{(n+m)!}{n!}$ $\displaystyle +\frac{(k+1+m)!}{(k+1)!}=-\frac{\Gamma (1+m) \Gamma (k+2)+m \Gamma (1+m) \Gamma (k+2)-\Gamma (k+3+m)}{(1+m) \Gamma (k+2)}$

    Knowing that $\displaystyle \Gamma(u)=(u-1)!$, we see that $\displaystyle \sum_{n=1}^k\frac{(n+m)!}{n!}$ $\displaystyle +\frac{(k+1+m)!}{(k+1)!}=-\frac{m!(k+1)!+m!m(k+1)!-(k+2+m)!}{(1+m)(k+1)!}$

    We can further simplify $\displaystyle -\frac{m!(k+1)!+m!m(k+1)!-(k+2+m)!}{(1+m)(k+1)!}=-\frac{(1+m)m!(k+1)!-(k+2+m)!}{(1+m)(k+1)!}$ $\displaystyle =-\frac{(m+1)!(k+1)!-(k+2+m)!}{(1+m)(k+1)!}$

    Since $\displaystyle \sum _{n=1}^k \frac{(n+m)!}{n!}=-\frac{\Gamma (1+m) \Gamma (1+k)+m \Gamma (1+m) \Gamma (1+k)-\Gamma (2+m+k)}{(1+m) \Gamma (1+k)}$ $\displaystyle =-\frac{m!k!+m!k!m-(k+1+m)!}{(1+m) k!}$

    Thus, we need to show that $\displaystyle -\frac{m!k!+m!k!m-(k+1+m)!}{(1+m) k!}+\frac{(k+1+m)!}{(k+1)!}$ $\displaystyle =\color{red}-\frac{(m+1)!(k+1)!-(k+2+m)!}{(1+m)(k+1)!}$

    Let's get the left side to look like the right side:

    $\displaystyle -\frac{m!k!+m!k!m-(k+1+m)!}{(1+m) k!}+\frac{(k+1+m)!}{(k+1)!}$ $\displaystyle =-\frac{(1+m)m!k!-(k+1+m)!}{(1+m)(1+k) k!}(k+1)+\frac{(1+m)(k+1+m)!}{(1+m)(k+1)k!}$

    $\displaystyle =\frac{-(m+1)!(k+1)!+(k+1)(k+1+m)!+(m+1)(k+1+m)!}{(m+1)(k+ 1)!}$ $\displaystyle =\frac{-(m+1)!(k+1)!+(k+2+m)(k+1+m)!}{(m+1)(k+1)!}$

    $\displaystyle =\frac{-(m+1)!(k+1)!+(k+2+m)!}{(m+1)(k+1)!}=\color{red}-\frac{(m+1)!(k+1)!-(k+2+m)!}{(m+1)(k+1)!}$

    This completes the inductive step.



    Does this make sense? Hopefully you can follow what I did...

    --Chris

    P.S. I wonder if anyone knows of a shorter way!?!?!?!

    Thank you for the answer. I've solved it using induction myself. But what if the question is "express $\displaystyle \sum _{n=1}^N \frac{(n+m)!}{n!}$ in terms of N and m
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  4. #4
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    Quote Originally Posted by fobos3 View Post

    Prove that $\displaystyle \sum _{n=1}^N \frac{(n+m)!}{n!}=-\frac{\Gamma (1+m) \Gamma (1+N)+m \Gamma (1+m) \Gamma (1+N)-\Gamma (2+m+N)}{(1+m) \Gamma (1+N)}$
    using the basic identity: $\displaystyle \binom{k-1}{\ell}=\binom{k}{\ell}-\binom{k-1}{\ell -1},$ we will have: $\displaystyle \sum_{n=1}^N \binom{n+m}{n}=\sum_{n=1}^N \left[\binom{n+m+1}{n}-\binom{n+m}{n-1} \right]. \ \ \ \ \ \ \ \ \ (1)$

    now the sum in the right hand side of (1) is a nice telescoping sum. thus (1) gives us: $\displaystyle \sum_{n=1}^N \binom{n+m}{n}= \binom{N+m+1}{N} - 1. \ \ \ \ \ \ \ (2)$

    finally multiplying both sides of (2) by $\displaystyle m!$ gives us: $\displaystyle \sum_{n=1}^N \frac{(n+m)!}{n!}=\left[\binom{N+m+1}{N}-1 \right]m!,$ which completes the proof because the

    nice looking right hand side of my identity is equal to the weird looking right hand side of your identity! $\displaystyle \Box$
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