# Does this series converge - for which a,b,c?

1. Mar 9, 2006

"Given a series whose general term is

$$\frac{a(a+1)...(a+n)b(b+1)...(b+n)}{(n+1)!c(c+1)...(c+n)}$$

prove that it converges if c>a+b, and a,b,c are strictly nonnegative"

I have tried all the tests I know (ratio, root, raabe's, abel), and they all failed. I don't know how to apply the integral test here, but I'm sure it will fail too.

Last edited: Mar 9, 2006
2. Mar 9, 2006

### NateTG

The ratio test worked for me.

3. Mar 9, 2006

I keep getting 1. How did you do it?

I get

$$\frac{(a+n+1)(b+n+1)}{(n+2)(c+n+1)}$$

which will result in 1 when taken to infinity.

4. Mar 9, 2006

### shmoe

I don't see how to use the ratio test either.

Do you know about the gamma function? You can write this in terms gammas, then use one of it's asymptotic relationships (Stirling's isn't the simplest choice here).

5. Mar 9, 2006

Even if I did, I doubt I would be allowed to use it. I'm beginning to think this can't converge at all, but it must.

6. Mar 10, 2006

### HallsofIvy

Staff Emeritus
It clearly will converge because of that factorial in the denominator.

7. Mar 10, 2006

### shmoe

I don't think this is an obvious thing, there are 2n+2 terms in both the numerator and denominator.

Try bounding your terms by taking the logarithm then grouping terms. Use basic bounds for the log, the goal in mind is to bound your entire term by a constant times something like n^{a+b-1-c}.

8. Mar 10, 2006

### benorin

This series ought to look farmiliar: define the Pochhammer symbol (a.k.a. the rising factorial) for nonnegative integers k by

$$(p)_k:=p(p+1)\cdots (p+k-1) = \frac{\Gamma (p+k)}{\Gamma (p)},(p)_0:=1$$

then your general term becomes

$$\frac{a(a+1)...(a+n)b(b+1)...(b+n)}{(n+1)!c(c+1).. .(c+n)}=\frac{(a)_{n+1}(b)_{n+1}}{(n+1)!(c)_{n+1}}$$

so it is evident that the series under consideration is the hypergeometric function

$$_2F_1 (a,b;c;z) = \sum_{n=0}^{\infty}\frac{(a)_{n}(b)_{n}}{(c)_{n}}\frac{z^n}{n!}$$

evaluated at z=1 and shifted a bit, so its

$$\sum_{n=0}^{\infty}\frac{(a)_{n+1}(b)_{n+1}}{(c)_{n+1}}\frac{1}{(n+1)!} = \sum_{n=1}^{\infty}\frac{(a)_{n}(b)_{n}}{(c)_{n}}\frac{1}{n!} = -1+\sum_{n=0}^{\infty}\frac{(a)_{n}(b)_{n}}{(c)_{n}}\frac{1}{n!} =-1 + _2F_1 (a,b;c;1)$$

Last edited: Mar 10, 2006
9. Mar 10, 2006

The method used is too advanced. There must be a more elementary proof.

10. Mar 10, 2006

### shmoe

Unless you already knew the relevant hypergeometric series converges, it wouldn't get you very far.

Did you try taking the log as I suggested? You won't need anything more advanced than a bound for things like log(1-x)

11. Mar 10, 2006

I don't really understand what you mean by taking the log. Do you mean taking the log of the general term, the entire series, or the ration of two consecutive terms?

Also, can someone verify Raabe's test? I did it again, and according to it, the series will diverge no matter what. There might be an error in my arithmetic.

12. Mar 10, 2006

### shmoe

of the terms:

$$\log\left(\frac{a(a+1)...(a+n)b(b+1)...(b+n)}{(n+1)!c(c+1).. .(c+n)}\right)$$

Though I missed your mention of Raabe's test, sorry, so you won't need to use my suggestion. Check Raabe's again, and post your result. You may have inverted the $$a_{n}/a_{n+1}$$term.

Last edited: Mar 10, 2006
13. Mar 10, 2006

Raabe's test was given to me as $$\frac{a_{n+1}}{a_n}\leq 1-\frac{\alpha}{n}, \alpha>1$$. Are you saying that it's actually $$\frac{a_{n}}{a_{n+1}}$$ instead?