How simplifying this ratio test for series?

[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.

 

[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)!

 

[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.

 

[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.

 

[frasifrasi]

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

 

[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.

 

[D H]

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 (2(n+1))!=(2n+2)(2n+1)(2n)![/tex]. Inserting these in the limit expression yields<br /> <br /> \lim_{n\to\infty}\frac {(n+1)n!(n+1)^2}{(2n+2)(2n+1)(2n)!}\frac {(2n)!}{n!n^2}<br /> <br /> Simplifying,<br /> <br /> \lim_{n\to\infty}\frac {(n+1)^3}{(2n+2)(2n+1)n^2}<br /> <br /> Most of the remaining terms cancel in the limit. For example,<br /> <br /> \lim_{n\to\infty}\frac{n+1}{n} = 1+\lim_{n\to\infty}\frac{1}{n} = 1

 

[rock.freak667]

rock.freak,

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

Yes it should...I made a mistake

 

[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.

 

[D H]

The only terms I canceled were n!/n! and (2n)!/(2n)![/tex]. I left the evaluation of<br /> <br /> \lim_{n\to\infty}\frac {(n+1)^3}{(2n+2)(2n+1)n^2}<br /> <br /> up to you, with a very strong hint in<br /> <br /> \lim_{n\to\infty}\frac{n+1}{n} = 1<br /> <br /> I guess the hint wasn&#039;t quite enough. Just drop the +1/+2 terms: change (n+1)^3, 2n+2, and 2n+1 to n^3, 2n, and 2n, respectively.

 

[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.

 

[D H]

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

 

[frasifrasi]

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

 

[D H]

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}

 

[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.

 

[D H]

I see you were just providing an example?

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

Thanks for the help!

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