How simplifying this ratio test for series?

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  • #1
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[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.
 

Answers and Replies

  • #2
rock.freak667
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(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
Gib Z
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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
Curious3141
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(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.
 
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  • #5
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I see, does anyone know a place I can look up these formulas?
 
  • #6
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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
D H
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I assume you are look at the convergence of

[tex]\sum_n \frac {n!n^2}{(2n)!}[/tex]

Applying the ratio test, you want to determine

[tex]\lim_{n\to\infty}\frac {(n+1)!(n+1)^2}{(2(n+1))!}\frac {(2n)!}{n!n^2}[/tex]

First expand the factorials: [itex](n+1)!=(n+1)n![/tex] and [itex](2(n+1))!=(2n+2)(2n+1)(2n)![/tex]. Inserting these in the limit expression yields

[tex]\lim_{n\to\infty}\frac {(n+1)n!(n+1)^2}{(2n+2)(2n+1)(2n)!}\frac {(2n)!}{n!n^2}[/tex]

Simplifying,

[tex]\lim_{n\to\infty}\frac {(n+1)^3}{(2n+2)(2n+1)n^2}[/tex]

Most of the remaining terms cancel in the limit. For example,

[tex]\lim_{n\to\infty}\frac{n+1}{n} = 1+\lim_{n\to\infty}\frac{1}{n} = 1[/tex]
 
  • #8
rock.freak667
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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
 
  • #9
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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
D H
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The only terms I canceled were [itex]n!/n![/itex] and [itex](2n)!/(2n)![/tex]. I left the evaluation of

[tex]\lim_{n\to\infty}\frac {(n+1)^3}{(2n+2)(2n+1)n^2}[/tex]

up to you, with a very strong hint in

[tex]\lim_{n\to\infty}\frac{n+1}{n} = 1[/tex]

I guess the hint wasn't quite enough. Just drop the +1/+2 terms: change [itex](n+1)^3[/itex], [itex]2n+2[/itex], and [itex]2n+1[/itex] to [itex]n^3[/itex], [itex]2n[/itex], and [itex]2n[/itex], respectively.
 
  • #11
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I see, so I would get [tex]\lim_{n\to\infty}\frac{n^3}{2n.2n.n^2} [/tex]

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
D H
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Yes, the ratio test limit is [itex]\lim_{n\to\infty}1/(4n)=0[/itex]. The series converges.
 
  • #13
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but, how/why did you get 1 in your previous post?
 
  • #14
D H
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That was not the same limit. That was

[tex]\lim_{n\to\infty}\frac{n+1}{n}[/tex]

which is indeed one but is also quite different from

[tex]\lim_{n\to\infty}\frac {(n+1)^3}{(2n+2)(2n+1)n^2}[/tex]
 
  • #15
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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.
 
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  • #16
D H
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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.
 

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