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

[Treadstone 71]

"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.

 

[NateTG]

The ratio test worked for me.

 

[Treadstone 71]

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.

 

[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).

 

[Treadstone 71]

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.

 

[HallsofIvy]

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

 

[shmoe]

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

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}.

 

[benorin]

"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.

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)

 

[Treadstone 71]

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

 

[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)

 

[Treadstone 71]

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.

 

[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.

 

[Treadstone 71]

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?

EDIT: I checked mathworld. It seems my professor gave me Raabe's test upside-down. It works now. Thanks.