# Proof of Divergence?

1. Jul 10, 2011

### talk2glenn

1. The problem statement, all variables and given/known data

Decide whether the series below is absolutely convergent, conditionally convergent, or divergent:

$$\sum_{1}^{\infty}(2n+3)!/(n!)^2$$

3. The attempt at a solution

By graphing the equation, I am confident that the series is divergent, but I don't know how to prove it. I cannot do the algebraic manipulation for a ratio test, assuming it is even possible, and none of the other tests seem applicable. Since it's apparently going to be divergent, I can't to a comparison test.

That leaves either a straight limit test, or a limit comparison test. Unfortunately, it looks to me like the limit converges to zero. Factorial is stronger than the power function, but how can I prove factorial squared is weaker than a larger factorial in the numerator?

Thanks :)

2. Jul 10, 2011

### Samuelb88

The series diverges by the ratio test. Can you show what you've done using the ratio test?

3. Jul 10, 2011

### talk2glenn

I can try....

$$\lim_{n\rightarrow\infty}(2n+5)!/((n+1)!)^2 * (n!)^2/(2n+3)!$$

If we expand and simplify [(n+1)!]^2, we get n!(n+1)n!(n+1)

Cancel the two n! in the numerator and denominator to get....

$$\lim_{n\rightarrow\infty}(2n+5)!/[(n+1)^2(2n+3)!]$$

I can't further simplify this expression, or pove that it is greater than 1.

Is this the right track?

4. Jul 10, 2011

### Samuelb88

So far so good.

Hint: $(2n+5)! = (2n+5)(2n+4)(2n+3)!$

5. Jul 10, 2011

### talk2glenn

Yeah I'm dumb...

So canceling the common factorials gives us....

$$\lim_{n\rightarrow\infty}(2n+5)(2n+4)/(n+1)(n+1)$$

Expand and approximate using leading coefficients to get...

$$\approx 4n^2/n^2$$

Which converges to 4/1 > 1 at the limit, so divergent. I think this is correct now.

Thank you so much!