Factorial gamma function

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November 4th 2008, 09:33 PM
fobos3

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

Any help will be appreciated
November 4th 2008, 10:38 PM
Chris L T521

Quote:

Originally Posted by fobos3

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

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:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

This completes the inductive step.

(Whew)

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

--Chris

P.S. I wonder if anyone knows of a shorter way!?!?!?! (Wondering)
November 4th 2008, 11:06 PM
fobos3

Quote:

Originally Posted by Chris L T521

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:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

This completes the inductive step.

(Whew)

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

--Chris

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

Thank you for the answer. I've solved it using induction myself. But what if the question is "express $\sum _{n=1}^N \frac{(n+m)!}{n!}$ in terms of N and m
November 5th 2008, 01:10 AM
NonCommAlg

Quote:

Originally Posted by fobos3

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

using the basic identity: $\binom{k-1}{\ell}=\binom{k}{\ell}-\binom{k-1}{\ell -1},$ we will have: $\sum_{n=1}^N \binom{n+m}{n}=\sum_{n=1}^N \left[\binom{n+m+1}{n}-\binom{n+m}{n-1} \right]. \ \ \ \ \ \ \ \ \ (1)$

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

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