limit

[alejandrito20]

Hello

The problem is

find the value of [math]\lambda[/math] for [math]\lim_{n \to \infty} \frac{a_{n+1}}{a_n} \nonumber\ [/math] < 1, where

 

[math] a_n= \frac{(\lambda^nn!)^2}{(2n+1)!} \nonumber\ [/math] con [math]\lambda >0 [/math]

 

I tried to do:

[math]\frac{a_{n+1}}{a_n}=\frac{(\lambda^{n+1}(n+1)!)^2}{(2n+3)!}\cdot \frac{(2n+1)!}{(\lambda^nn!)^2}[/math]

 

[math]=( \frac{\lambda^{n+1}}{\lambda^n}\frac{(n+1)!}{n!})^2\frac{(2n+1)!}{(2n+3)!} [/math]

 

[math]\frac{(2n+1)!}{(2n+3)!}= \frac{3 \cdot 5 \cdot 7........(2n+1)}{5 \cdot 7 ........(2n+1) \cdot (2n+3)}[/math]

 

[math]= ( \lambda (n+1))^2\frac{3}{(2n+3)} [/math]

 

but the Answer is [math]\lambda \in {0,2}[/math]

Edited May 14, 2011 by alejandrito20

[DJBruce]

Look at your third line of equations; you simplified your factorials incorrectly.

 

[math]\frac{(2n+1)!}{(2n+3)!}= \frac{(2n+1)(2n)...}{(2n+3)(2n+2)(2n+1)...1}[/math]

 

If you correct this you should notice that it works out correctly. However, don't forget to check your endpoints.

Edited May 14, 2011 by DJBruce

[DrRocket]

Hello

The problem is

find the value of [math]\lambda[/math] for [math]\lim_{n \to \infty} \frac{a_{n+1}}{a_n} \nonumber\ [/math] < 1, where

 

[math] a_n= \frac{(\lambda^nn!)^2}{(2n+1)!} \nonumber\ [/math] con [math]\lambda >0 [/math]

 

I tried to do:

[math]\frac{a_{n+1}}{a_n}=\frac{(\lambda^{n+1}(n+1)!)^2}{(2n+3)!}\cdot \frac{(2n+1)!}{(\lambda^nn!)^2}[/math]

 

[math]=( \frac{\lambda^{n+1}}{\lambda^n}\frac{(n+1)!}{n!})^2\frac{(2n+1)!}{(2n+3)!} [/math]

 

[math]\frac{(2n+1)!}{(2n+3)!}= \frac{3 \cdot 5 \cdot 7........(2n+1)}{5 \cdot 7 ........(2n+1) \cdot (2n+3)}[/math]

 

[math]= ( \lambda (n+1))^2\frac{3}{(2n+3)} [/math]

 

but the Answer is [math]\lambda \in {0,2}[/math]

 

 

 

If [math] a_n= \frac{(\lambda^nn!)^2}{(2n+1)!} \nonumber\ [/math]

 

Then

 

[math]\frac{a_{n+1}}{a_n}=\frac{(\lambda^{n+1}(n+1)!)^2}{(2n+3)!}\cdot \frac{(2n+1)!}{(\lambda^nn!)^2}[/math]

 

[math]= (\frac{\lambda^{n+1}}{\lambda^n}\frac{(n+1)!}{n!})^2\frac{(2n+1)!}{(2n+3)!} [/math]

 

[math]= (\frac{\lambda}{1}\frac{(n+1)}{1})^2\frac{(1)}{(2n+3)(2n+2)} [/math]

 

[math]= \frac{\lambda^2 (n^2 +2n +1)}{4n^2 +10n +6)} [/math]

 

[math] = \frac {\lambda ^2}{4} \frac {n^2 +2n +1} {n^2 + \frac{5n}{2} + \frac {3}{4}}[/math] [math] \rightarrow \frac {\lambda ^2}{4}[/math] as [math] n \rightarrow \infty[/math]

[alejandrito20]

thanks :=)