# partial sum involving factorials

Here is an interesting series I ran across.

It is a binomial-type identity.

$\displaystyle \sum_{k=0}^{n}\frac{(2n-k)!\cdot 2^{k}}{(n-k)!}=4^{n}\cdot n!$

I tried all sorts of playing around, but could not get it to work out.

This works out the same as $\displaystyle 2^{n}\prod_{k=1}^{n}2k=2^{n}\cdot 2^{n}\cdot n!=4^{n}\cdot n!$

I tried equating these somehow, but I could not get it. I even wrote out the series.

There were cancellations, but it did not look like the product of the even numbers.

$\displaystyle \frac{(2n)!}{n!}+\frac{(2n-1)!\cdot 2}{(n-1)!}+\frac{(2n-2)!2^{2}}{(n-2)!}+\cdot\cdot\cdot +n!\cdot 2^{n}=4^{n}\cdot n!$.

How can the closed form be derived from this?. I bet I am just being thick. I see the last term is nearly the result except for being multiplied by $2^{n}$. I see if the factorials are written out, $2n(2n-1)(2n-2)(2n-3)\dots$ for example, then 2's factor out of $2n, \;\ 2n-2$ (even terms) in the numerator.

There is even a general form I ran through Maple. It actually gave a closed from for it as well, but I would have no idea how to derive it.

$\displaystyle \sum_{k=0}^{n}\frac{(2n-k)!\cdot 2^{k}\cdot (k+m)!}{(n-k)!\cdot k!}$.

In the above case, m=0. But, apparently there is a closed form for $m\in \mathbb{N}$ as well.

Maple gave the solution in terms of Gamma: $\displaystyle \frac{\Gamma(1+m)4^{n}\Gamma(n+1+\frac{m}{2})}{\Gamma(1+\frac{m}{2})}$

Would anyone have an idea how to proceed with this?. Perhaps writing it in terms of Gamma and using some identities?. Thanks very much.

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Your "general form" doesn't make sense if $m$ is a positive integer, because you'll get factorials of negative integers: $(k-m)!$. Perhaps you mean $(m-k)!$ instead? –  Robert Israel Dec 30 '11 at 18:25
If you flip the summation order of your first sum, it becomes $$2^n n!\sum\limits_{k=0}^n \binom{n+k}{k}\frac1{2^k}$$ –  Guess who it is. Dec 30 '11 at 18:47

\begin{align} \sum_{k=0}^{n}\frac{(2n-k)!}{(n-k)!}2^k\tag{1} &=n!\sum_{k=0}^n\binom{2n-k}{n-k}\sum_{j=0}^k\binom{k}{j}\\\tag{2} &=n!\sum_{k=0}^n\sum_{j=0}^{n-k}\binom{n+k}{k}\binom{n-k}{j}\\\tag{3} &=n!\sum_{j=0}^n\sum_{k=0}^{n-j}\binom{n+k}{n}\binom{n-k}{j}\\\tag{4} &=n!\sum_{j=0}^n\binom{2n+1}{n+j+1}\\\tag{5} &=n!\;2^{2n} \end{align}

1. rewrite $2^k$

2. $k\mapsto n-k$

3. change order of summation and $\binom{n}{k}=\binom{n}{n-k}$

4. $\sum_k\binom{n-k}{i}\binom{m+k}{j}=\binom{n+m+1}{i+j+1}$

5. Split $\sum_j\binom{2n+1}{j}=2^{2n+1}$ in half

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A virtuoso use of binomial coefficient identities! –  Mike Spivey Dec 31 '11 at 3:54

Here's a combinatorial proof for J.M.'s reformulation (after dividing out $n!$):

$$\sum\limits_{k=0}^n \binom{n+k}{k} 2^{n-k}= 4^n.$$

Suppose you flip coins until you obtain either $n+1$ heads or $n+1$ tails. After either heads or tails "wins" you keep flipping until you have a total of $2n+1$ coin flips. The two sides count the number of ways for heads to win.

For the left side: Condition on the number of tails $k$ obtained before head $n+1$. There are $\binom{n+k}{k}$ ways to choose the positions at which these $k$ tails occurred from the $n+k$ total options, and then $2^{n-k}$ possibilities for the remaining flips after head $n+1$. Summing up yields the left side.

For the right side: Heads wins on half of the total number of sequences; i.e., $\frac{1}{2}(2^{2n+1}) = 4^n$.

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Nice! More direct than my detour through Concrete Mathematics. –  Byron Schmuland Dec 30 '11 at 20:05

Dividing out $2^n\cdot n!$ and changing variables, your equation follows from (5.20) on page 167 of Concrete Mathematics by Graham, Knuth, and Patashnik. If you don't have this reference, I can add more details later.

$$\sum_{k\leq m}{m+k\choose k}2^{-k}=2^m\qquad\text{integer}\quad m\geq 0.\qquad\qquad (5.20)$$

Just for fun, here is a probabilistic proof of (5.20). Suppose we toss a fair coin until we see either $m+1$ heads or $m+1$ tails. Let $J$ be the number of trials prior to the final toss, and consider $\mathbb{P}(J=j)$ for $m\leq j\leq 2m$. We get $J=j$ by either $m$ heads in the first $j$ tosses followed by a head, or $m$ tails in the first $j$ tosses followed by a tail. By symmetry, we find $$\mathbb{P}(J=j)=2{j\choose m}\left({1\over 2}\right)^j\left({1\over 2}\right) ={j\choose m}\left({1\over 2}\right)^j.$$

Since the total probability adds to one, we get $$\sum_{m\leq j\leq 2m} {j\choose m}\left({1\over 2}\right)^j=1.$$ Putting $k=j-m$ in this sum and multiplying by $2^m$ gives (5.20).

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Interesting reference (GKP) thanks! Page 247 they propose to prove that $\sum_{k\gt m}{m+k\choose k}2^{-k}=2^m$ using the additional fact that $(1-x)^{-1-n} = \sum^{\infty}_{k=0} \binom{k+n}{n} x^k$ (see (mathoverflow.net/questions/84523/an-asymptotic-question)) –  Raymond Manzoni Dec 30 '11 at 19:21
@Raymond Thanks for the link! –  Byron Schmuland Dec 30 '11 at 19:59
I am so sorry. I had a typo in the original post. I even looked it over to make sure and still missed it. That should have been $(k+m)!$. I made some edits. Of course, that changed the closed form which can be written with factorials instead of Gamma: $\displaystyle \frac{m!4^{n}(n+\frac{m}{2})!}{(\frac{m}{2})!}$. Again, my apologies and thanks for all the responses. No, I do not have Concrete Mathematics, but I think I can obtain it. I will look into it. Thanks –  Cody Dec 30 '11 at 22:22
@Cody: you definitely should; it's a nice book to have! –  Guess who it is. Dec 31 '11 at 0:35

This identity can be re-written as $$\sum_{k=0}^n {2n-k \choose n-k} 2^k = 4^n.$$

Start from $${2n-k \choose n-k} = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{(1+z)^{2n-k}}{z^{n-k+1}} \; dz.$$

This yields for the sum $$\frac{1}{2\pi i} \int_{|z|=\epsilon} \sum_{k=0}^n \frac{(1+z)^{2n-k}}{z^{n-k+1}} 2^k \; dz \\ = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{(1+z)^{2n}}{z^{n+1}} \sum_{k=0}^n \frac{(2z)^{k}}{(1+z)^k} \; dz.$$

We can extend the sum to infinity because when $n-k+1 \le 0$ or $k \ge n+1$ the integrand of the defining integral of the binomial coefficient is an entire function and the integral is zero. This yields

$$\frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{(1+z)^{2n}}{z^{n+1}} \sum_{k=0}^\infty \frac{(2z)^{k}}{(1+z)^k} \; dz \\ = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{(1+z)^{2n}}{z^{n+1}} \frac{1}{1-2z/(1+z)} \; dz \\ = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{(1+z)^{2n+1}}{z^{n+1}} \frac{1}{1-z} \; dz.$$

Thus the value of the integral is given by $$[z^n] \frac{1}{1-z} (1+z)^{2n+1} = \sum_{q=0}^n {2n+1\choose q} = \frac{1}{2} 2^{2n+1} = 4^n.$$

A trace as to when this method appeared on MSE and by whom starts at this MSE link.

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