# How simplifying this ratio test for series?

1. Oct 31, 2007

### frasifrasi

[SOLVED] How simplifying this ratio test for series?

Basically, we are asked if (n!n^2)/(2n)! converges absolutely...

I got to the point where

lim as n --> infinity of

[(n+1)!(n+1)^2]/(2(n+1))! X (2n)!/n!n^2

by the ratio test.

But I don't know how to manipulate the factorials from here on...could anyone help?

Thank you.

2. Oct 31, 2007

### rock.freak667

(n+1)!=(n+1)n! which should cancel with the n!
2(n+1))! =(2n+2)!=(2n+2)(2n-1)!
(2n)!=2n(2n-1)!

3. Nov 1, 2007

### Gib Z

You could replace the factorials with their respective Stirling approximations, which is asymptotic to the factorial function, ie as n grows large the quotient of the approximation and the factorial approaches 1.

4. Nov 1, 2007

### Curious3141

(My earlier idea was wrong because I compared with a divergent series, but adapted slightly and if you're allowed to assume the convergence of zeta(2), the following might do the trick simply).

For large n, the denominator (2n)! = (n+n)! >> (n+4)! = n!(n+1)(n+2)(n+3)(n+4) > (n!)n^4

Cancel off the numerator and you're left with terms that are strictly less than 1/n^2 for large n. By the limit comparison test with zeta(2), this converges.

Last edited: Nov 1, 2007
5. Nov 1, 2007

### frasifrasi

I see, does anyone know a place I can look up these formulas?

6. Nov 1, 2007

### frasifrasi

rock.freak,

for the (2n+2)!, shouldn't it be (2n+2)(2n+1)! ... because you keep subtracting 1??

And can anyone please evaluate this series...I am getting stuck at the end with

lim as n --> infinity of (n+1)^2/n.

Thank you.

7. Nov 1, 2007

### D H

Staff Emeritus
I assume you are look at the convergence of

$$\sum_n \frac {n!n^2}{(2n)!}$$

Applying the ratio test, you want to determine

$$\lim_{n\to\infty}\frac {(n+1)!(n+1)^2}{(2(n+1))!}\frac {(2n)!}{n!n^2}$$

First expand the factorials: $(n+1)!=(n+1)n![/tex] and [itex](2(n+1))!=(2n+2)(2n+1)(2n)![/tex]. Inserting these in the limit expression yields $$\lim_{n\to\infty}\frac {(n+1)n!(n+1)^2}{(2n+2)(2n+1)(2n)!}\frac {(2n)!}{n!n^2}$$ Simplifying, $$\lim_{n\to\infty}\frac {(n+1)^3}{(2n+2)(2n+1)n^2}$$ Most of the remaining terms cancel in the limit. For example, $$\lim_{n\to\infty}\frac{n+1}{n} = 1+\lim_{n\to\infty}\frac{1}{n} = 1$$ 8. Nov 1, 2007 ### rock.freak667 Yes it should...I made a mistake 9. Nov 1, 2007 ### frasifrasi I still don't see how you evaluate that last (n+1)^3 limit, though. DH, did you expand the terms or cancel something out? Thank you everyone for the great help. 10. Nov 1, 2007 ### D H Staff Emeritus The only terms I canceled were [itex]n!/n!$ and $(2n)!/(2n)![/tex]. I left the evaluation of $$\lim_{n\to\infty}\frac {(n+1)^3}{(2n+2)(2n+1)n^2}$$ up to you, with a very strong hint in $$\lim_{n\to\infty}\frac{n+1}{n} = 1$$ I guess the hint wasn't quite enough. Just drop the +1/+2 terms: change [itex](n+1)^3$, $2n+2$, and $2n+1$ to $n^3$, $2n$, and $2n$, respectively.

11. Nov 1, 2007

### frasifrasi

I see, so I would get $$\lim_{n\to\infty}\frac{n^3}{2n.2n.n^2}$$

SO, here I cancel the n^3 with the n^2, correct, getting n/(2n)(2n) -- Don't I get zero??

A little more effort on your part and I am on my way to grasping this. Thanks.

12. Nov 1, 2007

### D H

Staff Emeritus
Yes, the ratio test limit is $\lim_{n\to\infty}1/(4n)=0$. The series converges.

13. Nov 1, 2007

### frasifrasi

but, how/why did you get 1 in your previous post?

14. Nov 1, 2007

### D H

Staff Emeritus
That was not the same limit. That was

$$\lim_{n\to\infty}\frac{n+1}{n}$$

which is indeed one but is also quite different from

$$\lim_{n\to\infty}\frac {(n+1)^3}{(2n+2)(2n+1)n^2}$$

15. Nov 1, 2007

### frasifrasi

I see you were just providing an example? that actually confused me a lot because i thought it was the next sterp.

Thanks for the help!

I guess what i get is 1/4n, which goes to zero.

Last edited: Nov 1, 2007
16. Nov 1, 2007

### D H

Staff Emeritus
That's why I said "Most of the remaining terms cancel in the limit. For example,"

You're welcome. I'm marking this as solved.