Proof that series diverges ? (Defn of factorial ?)

[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 ideas 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 that's wrong can't figure out why though ?

 

[DryRun]

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.

 

[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)!

 

[Infinitum]

However, i have a doubt about (4n)! = 4(n!) or 4!n! ?? In my calculation above, i used the first one.

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.

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 ?)

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

 

[DryRun]

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.

 

[sid9221]

Just to clarify the factorial thing.

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

 

[SammyS]

Just to clarify the factorial thing.

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

Yes, that's correct.