# Difficult Double Series

1. Aug 29, 2007

I have tried well over 30 cumulative hours trying to evaluate this double series:

$$S=\sum_{k=0}^{m}\sum_{j=0}^{k+m-1}(-1)^{k}{m \choose k}\frac{[2(k+m)]!}{(k+m)!^{2}}\frac{(k-j+m)!^{2}}{(k-j+m)[2(k-j+m)]!}\frac{1}{2^{k+j+m+1}},$$

for some integer $$m>0$$. I have simplified (or maybe complicated) the sum to

$$S=\sum_{k=0}^{m}\sum_{j=0}^{k+m-1}(-1)^{k}\frac{m!}{k!(m-k)!}\frac{(k+m-1-j)!}{(k+m)!}\frac{(k+m-1/2)!}{(k+m-1/2-j)!}\frac{1}{2^{k-j+m+1}},$$

which may be easier to work with (hopefully it is correct).

I conjecture that $$S=0$$ from calculating $$S$$ with 50 consecutive values of $$m$$. And assuming Mathematica knows what it's doing, $$m$$ has to equal zero from earlier results; results preceding the derivation of this double sum.

I am wondering if anyone knows an easy way to evaluate this sum. :)

Last edited: Aug 30, 2007
2. Aug 30, 2007

### Gib Z

I'm on it!

3. Aug 30, 2007

### jostpuur

What do notations [n]! and (n+1/2)! mean?

4. Aug 30, 2007

jostpuur: $$z!$$ reads "$$z$$ factorial." If $$z \in \mathbb{N}$$, then $$z!=z(z-1)(z-2)\cdots 2\cdot 1$$. For $$z=0$$, then $$0!=1$$.

This idea can also be extended to all real numbers (with a restriction). If $$z \in \mathbb{R}$$, then $$z!=\Gamma(z+1)$$, where $$\Gamma(\cdot)$$ is the Gamma function. Notice that $$z$$ cannot be a non-positive integer here, because at those values, vertical asymptotes exist.

Everyone else: I have taken many approaches into evaluating this double series. For example, I have fiddled with hypergeometric functions a bit, tried proving that the sum of all positive terms equals the sum of all negative terms, tried manipulating the summand (heavily), made many substitutions...and I cannot get it. What makes this double series difficult is that $$|S|$$, for some integer $$m>0$$ somewhat behaves like a non-symmetrical binomial distribution. For those who are unfamiliar, it is a discrete statistical distribution whose points approximately lie on a translated, continuous normal distribution. But the problem is that these points are not symmetrical. In other words, there are no pairwise cancellations in my series, which I initially suspected. To be a little more specific, let

$$s(k) = \frac{(-1)^{k}}{2^{k+m+1}}{m \choose k}\frac{[2(k+m)]!}{(k+m)!^{2}}\sigma(k),$$

where

$$\sigma (k)=\sum_{j=0}^{k+m-1}\frac{(k-j+m)!^{2}}{(k-j+m)[2(k-j+m)]!}\frac{1}{2^{j}}.$$

There are no two integers $$0 \le k \le m$$ (call them $$k_1$$ and $$k_2$$) which $$s(k_1)+s(k_2)=0$$. I have no proof of this, but graphical and numerical evidence supports this.

This is a problem assuming my initial conjecture is correct: that

$$S = \sum^m_{k=0}s(k) = 0,$$

which it seems to be for $$1 \le m \le 50$$.

I hope some of this information helped!

Last edited: Aug 30, 2007
5. Aug 30, 2007

### jostpuur

I know factorial, but does the [n] have some meaning, or is it just [n]=n?

6. Aug 30, 2007

The square brackets have no meaning (other than to group expressions); they are just used in the traditional fashion to avoid many parentheses. So, yes, $$[z]=(z)=z$$.

Last edited: Aug 30, 2007
7. Aug 31, 2007

### quidamschwarz

It might help to view some of your expressions as binomials!?

$$\frac{[2(k+m)]!}{(k+m)!^{2}} = {2(k+m) \choose k+m}$$

and

$$\frac{(k-j+m)!^{2}}{[2(k-j+m)]!} = {2(k-j+m) \choose k-j+m}^{-1}\;.$$

So we have

$$S=\sum_{k=0}^{m}\sum_{j=0}^{m+k-1}(-1)^{k}{m \choose k} {2(m+k) \choose m+k} {2(m+k-j) \choose m+k-j}^{-1} \frac{1}{2^{m+k+j+1}\,(m+k-j)}\;,$$

where I also commuted some $k$ and $m$. Maybe someone knows some fancy binomal-relations...

8. Aug 31, 2007

### Gib Z

I have a truly marvelous solution. I would type it up for you if the character limit of the post weren't so small.... =[

9. Aug 31, 2007

### NateTG

Have you looked at:
S(n)-S(n-1)

(No clue if this will give you an easy answer.)

10. Aug 31, 2007

I have turned the whole thing...or at least most of the sum into binomial coefficients. And I looked for hours upon hours for some nice relations. But I could get to nothing. I even got an expression without any inverse (multiplicative) binomial coefficients (if I did my algebra correctly). Maybe someone else can come up with a novel idea, or use with binomial coefficients.

11. Aug 31, 2007

### Haelfix

Quad, what you have there is a complete mess and I suspect will take anyone here quite awhile to prove a closed form answer (if it exists).

My question is the following. Where does this come from? It might be easier to go back a couple steps and see if you can run approximations before hand or use a different method a priori. If the solution really is something simple, its usually the case that you can spot it quite a bit earlier.

Last edited: Aug 31, 2007
12. Aug 31, 2007

I was evaluating the integral

$$\frac{1}{2^n}\int^\infty_0 \mathrm{sech}^n\, x\, dx$$

without any information (such as reduction formulae) beforehand. I needed an original idea for a project I am working on.

If anyone wants to help me personally, I don't mind giving my notes (which contains the derivation and all that of this sum, as well as my proof for the integral) for review, but I do not want them public at the moment.

I am a high school student, so my knowledge in math is quite low. I don't know very many advanced techniques for such problems.

Last edited: Aug 31, 2007
13. Aug 31, 2007

### Gib Z

Aww come on mate! Would have been much easier to start like THAT!!

$$I= \frac{1}{2^n} \int^{\infty}_0 \frac{2^n}{(e^x + e^{-x})^n} dx = \int^{\infty}_0 (e^x + e^{-x})^{-n} dx$$.

Post number 4 tells me you don't mind dealing with hyper geometric functions, so heres the anti derivative I get:

$$\frac{(1+e^{2x})^n}{ n(e^x + e^{-x})^n} {_2F_1} \left( \frac{n}{2}, n ; \frac{n}{2} + 1 ; -e^{2x})$$

In case you don't like the hypergeometric function, in series form:
$$_2F_1 (a, b; c; d) = \sum_{n=0}^{\infty} \frac{ (a)_n (b)_n }{ (c)_n} \frac{d^n}{n!} = 1+ \frac{ab}{1!c}d + \frac{a(a+1)b(b+1)}{2!c(c+1)}d^2...$$

where (x)_n is the Pochhammer Symbol;
$$(x)_n = \frac{ \Gamma (x+n)}{\Gamma (x)} = x(x+1)(x+2)...(x+n-1)$$.

Have fun evaluating!!

Last edited: Aug 31, 2007
14. Aug 31, 2007

I have that antiderivative in my notes. But, I don't know how to derive it. Moreover, that is not the antiderivative I want. I found an equivalent form which might lead to something interesting. ;)

Last edited: Aug 31, 2007
15. Sep 1, 2007

### Gib Z

It can easily be derived be finding the taylor series of the integrand and integrating those term by term. Seeing as you are using the anti derivative for the purpose of evaluating an integral, any form should be sufficient should it not?

16. Sep 1, 2007