- #1
alejandrito29
- 150
- 0
Hello
The problem is
find the value of [tex]\lambda[/tex] for [tex]\lim_{n \to \infty} \frac{a_{n+1}}{a_n} \nonumber\ [/tex] < 1, where
[tex] a_n= \frac{(\lambda^nn!)^2}{(2n+1)!} \nonumber\ [/tex] con [tex]\lambda >0 [/tex]
I tried to do:
[tex]\frac{a_{n+1}}{a_n}=\frac{(\lambda^{n+1}(n+1)!)^2}{(2n+3)!}\cdot \frac{(2n+1)!}{(\lambda^nn!)^2}[/tex]
[tex]=( \frac{\lambda^{n+1}}{\lambda^n}\frac{(n+1)!}{n!})^2\frac{(2n+1)!}{(2n+3)!} [/tex]
[tex]\frac{(2n+1)!}{(2n+3)!}= \frac{3 \cdot 5 \cdot 7...(2n+1)}{5 \cdot 7 ...(2n+1) \cdot (2n+3)}[/tex]
[tex]= ( \lambda (n+1))^2\frac{3}{(2n+3)} [/tex]
but the Answer is [tex]\lambda \in {0,2}[/tex]
The problem is
find the value of [tex]\lambda[/tex] for [tex]\lim_{n \to \infty} \frac{a_{n+1}}{a_n} \nonumber\ [/tex] < 1, where
[tex] a_n= \frac{(\lambda^nn!)^2}{(2n+1)!} \nonumber\ [/tex] con [tex]\lambda >0 [/tex]
I tried to do:
[tex]\frac{a_{n+1}}{a_n}=\frac{(\lambda^{n+1}(n+1)!)^2}{(2n+3)!}\cdot \frac{(2n+1)!}{(\lambda^nn!)^2}[/tex]
[tex]=( \frac{\lambda^{n+1}}{\lambda^n}\frac{(n+1)!}{n!})^2\frac{(2n+1)!}{(2n+3)!} [/tex]
[tex]\frac{(2n+1)!}{(2n+3)!}= \frac{3 \cdot 5 \cdot 7...(2n+1)}{5 \cdot 7 ...(2n+1) \cdot (2n+3)}[/tex]
[tex]= ( \lambda (n+1))^2\frac{3}{(2n+3)} [/tex]
but the Answer is [tex]\lambda \in {0,2}[/tex]