Proof that series diverges ? (Defn of factorial ?)

In summary, the conversation discusses the ratio test and its application to a specific series. It is determined that the series diverges, despite the fact that the ratio goes to infinity. There is also a clarification on the meaning of (4n)! and how it differs from 4(n!) and 4!n!. The conversation ends with a final clarification on the correct way to factorize the expression for the limit of L.
  • #1
sid9221
111
0
[tex] Ʃ \frac{(4n)!}{n!(2n)!} [/tex]

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 ?
 
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  • #2
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
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
sharks said:
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. :redface:

[itex](4n)! = (4n)(4n-1)(4n-2)...(2)(1)[/itex]

where as,

[itex]4(n!) = 4(n)(n-1)(n-2)...(2)(1)[/itex]

and

[itex]4!n! = 24(n)(n-1)(n-2)....(2)(1)[/itex]

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.
 
  • #5
Thanks for the clarification. I must have been intimidated by all those factorials. :redface:

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

For the OP, here is the final result to find the limit of L:
[tex]\frac{4(16n^2+16n+3)}{n+1}[/tex]Then, just factorize n out of the numerator and denominator, and the limit becomes clear.
 
  • #6
Just to clarify the factorial thing.

(4(n+1))! = (4n+4)! = (4n+4)(4n+3)(4n+2)(4n+1)(4n)! right ??
 
  • #7
sid9221 said:
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.
 

Related to Proof that series diverges ? (Defn of factorial ?)

1. What is the definition of a divergent series?

A divergent series is a series in which the sum of its terms approaches infinity as the number of terms increases.

2. How can you prove that a series is divergent?

One way to prove that a series is divergent is by showing that the terms of the series do not approach zero as the number of terms increases, which is a necessary condition for convergence.

3. What is the definition of a factorial?

A factorial is a mathematical function represented by the symbol "!" that multiplies all positive integers less than or equal to a given number. For example, 5! = 5 x 4 x 3 x 2 x 1 = 120.

4. How is the definition of a factorial related to the divergence of a series?

The definition of a factorial is related to the divergence of a series because some series involve the factorial function, such as the series for e (the base of the natural logarithm), which is given by the sum of 1/n! for n = 0 to infinity. This series diverges, meaning that the terms of the series do not approach zero as the number of terms increases.

5. Can a series with a factorial in its terms ever converge?

Yes, a series with a factorial in its terms can converge if the terms of the series approach zero fast enough to counteract the growth of the factorial function. An example is the series for the sine function, which is given by the sum of (-1)^n / (2n+1)! for n = 0 to infinity. This series converges to a finite value, despite the presence of the factorial function in its terms.

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