# HelloThe problem isfind the value of $$\lambda$$ for

## Main Question or Discussion Point

Hello
The problem is
find the value of $$\lambda$$ for $$\lim_{n \to \infty} \frac{a_{n+1}}{a_n} \nonumber\$$ < 1, where

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

I tried to do:
$$\frac{a_{n+1}}{a_n}=\frac{(\lambda^{n+1}(n+1)!)^2}{(2n+3)!}\cdot \frac{(2n+1)!}{(\lambda^nn!)^2}$$

$$=( \frac{\lambda^{n+1}}{\lambda^n}\frac{(n+1)!}{n!})^2\frac{(2n+1)!}{(2n+3)!}$$

$$\frac{(2n+1)!}{(2n+3)!}= \frac{3 \cdot 5 \cdot 7........(2n+1)}{5 \cdot 7 ........(2n+1) \cdot (2n+3)}$$

$$= ( \lambda (n+1))^2\frac{3}{(2n+3)}$$

but the Answer is $$\lambda \in {0,2}$$

You write

$$(2n+1)!=3\cdot 5\cdot ...\cdot (2n+1)$$

but this is not true. The factorial sign demands you to multiplicate ALL numbers under 2n+1. So the correct formula is

$$(2n+1)=1\cdot 2\cdot 3\cdot 4\cdot ...\cdot (2n+1)$$

Thesame remark applies to 2n+3.

It's not because you take the factorial of 2n+1 that you can only multiplicate the odd numbers! In fact, 2n+1 is simply a number, for example if n=3, then 2n+1 is 7, and (2n+1)!=7!=7.6.5.4.3.2.1.

very thanks