# Proof that series diverges ? (Defn of factorial ?)

1. May 22, 2012

### sid9221

$$Ʃ \frac{(4n)!}{n!(2n)!}$$

I tried the ratio test, but if my solution has a limit which goes to infinity. Hence the test is inconclusive. (Or does L->inf imply divergence as L>1 ?)

Any idea's what other test might be useful ?

PS: I have used the idea that (2(n+1))! = (2n+2)! = (2n+2)(2n+1)(2n)!

Wolfram say's thats wrong can't figure out why though ?

Last edited: May 22, 2012
2. May 22, 2012

### sharks

I get the ratio test as: 1/(n+1) and the series converges, as L = 0 < 1.
However, i have a doubt about (4n)! = 4(n!) or 4!n! ?? In my calculation above, i used the first one.

3. May 22, 2012

### tamtam402

The series diverges. The fact that the ratio goes to infinity doesn't change anything, the only case where the ratio test is inconclusive is if the ratio is equal to 1.

sharks, (4n)! doesn't mean 4(n!), so when you use the ratio test it becomes [4(n+1)]! = (4n+4)! = (4n+4)(4n+3)(4n+2)(4n+1)(4n)!

4. May 22, 2012

### Infinitum

None of the above.

$(4n)! = (4n)(4n-1)(4n-2).........(2)(1)$

where as,

$4(n!) = 4(n)(n-1)(n-2)...........(2)(1)$

and

$4!n! = 24(n)(n-1)(n-2).............(2)(1)$

They are definitely not the same.

As tamtam402 said, since the limit goes to positive infinity, it implies the series diverges.

5. May 22, 2012

### sharks

Thanks for the clarification. I must have been intimidated by all those factorials.

L=∞ > 1, therefore the series diverges by the ratio test.

For the OP, here is the final result to find the limit of L:
$$\frac{4(16n^2+16n+3)}{n+1}$$Then, just factorize n out of the numerator and denominator, and the limit becomes clear.

6. May 22, 2012

### sid9221

Just to clarify the factorial thing.

(4(n+1))! = (4n+4)! = (4n+4)(4n+3)(4n+2)(4n+1)(4n)! right ??

7. May 22, 2012

### SammyS

Staff Emeritus
Yes, that's correct.