I need help about hypergeometric series

**miss_lolitta** · April 9th 2006, 03:26 PM

The hyper geometric series :the infinite series

[1+ (ab/1!c)x+[a(a+1)b(b+1)/(2!c(c+1)]x^2+{[a(a+1)(a+2)b(b+1)(b+2)]/[3!c(C+1)(c+1)(c+2)]}x^3+....]

Where a,b and c are neither 0 nor negative integers,,

Please can anyone help me to prove by explaining that the hyper geometric series is converges??!

any help would be appreciated ,thank you

**Jameson** · April 9th 2006, 04:07 PM

There are a few ways to write this, but try looking at it like this.

$S_{n}=1+\sum_{n=0}^{\infty}\frac{(a+n)!(b+n)!x^{n+ 1}}{(n+1)!(c+n)!}$

Whew. And I would definitely take the ratio test from here.

**miss_lolitta** · April 9th 2006, 09:26 PM

Thanks Jameson

but if you do'nt have any objection,can you give me the complete answer

I'm confused

and not sure if I can do this or no

Please Jameson..

thanks for any responses.

**Jameson** · April 11th 2006, 03:19 PM

Ok. I'll try to Tex this all out.

Let $b_{n}=\frac{a_{n+1}}{a_n}$ or $a_{n+1}*\frac{1}{a_n}$

So $b_n=\frac{(a+n+1)!(b+n+1)!x^{n+2}}{(n+2)!(c+n+1)!} *\frac{(n+1)!(c+n)!}{a+n)!(b+n)!x^{n+1}}$

All I did was substitute (n+1) for in, then divide that expression by a_n. But to simplify, I just flipped a_n and multiplied. Now for some cancellation.

$b_n=\left(\frac{(a+n+1)(a+n)!(b+n+1)(b+n)!x^{n+1}* x}{(n+2)(n+1)!(c+n+1)(c+n)!}\right)$ $\left(\frac{(n+1)!(c+n)!}{(a+n)!(b+n)!x^{n+1}} \right)$

For some reason, that Latex won't make that last bracket! Anyway, cancel some terms out and you'll get.

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

Now you have your common ratio! So take the limit as $n\rightarrow\infty$ to see how the terms trend torwards infinity, and you should be able to show that this converges for all x.

Hopefully that makes sense.

**miss_lolitta** · April 14th 2006, 02:38 PM

I try to take limit as $<br /> n\rightarrow\infty<br />$ I get $<br /> \lim b_n=x<br />$

And conclude that the series diverges for x>1 , converges for x<1 and test failed at x=1 .

We can use Gauss' test to see series behavior ..

But I have one question :
Is it necessary that we test series behavior at x=-1 or no??!

thanks

**Jameson** · April 16th 2006, 06:05 AM

Yes. Sorry. The series converges for $|x|<1$ and you must test the endpoints of 1 and -1 for convergence.

**miss_lolitta** · April 16th 2006, 11:15 AM

but that's very long & hard

anyway..

thanks so much for Jameson

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