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, 201114 yr by alejandrito20

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, 201114 yr by DJBruce

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]

thanks :=)