Proof of Divergence for Series (2n+3)!/(n!)^2 - Limit Test or Comparison Test?

[talk2glenn]

Homework Statement

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

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

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 :)

 

[Samuelb88]

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

 

[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?

 

[Samuelb88]

So far so good.

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

 

[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!