(((-1)^n)n!)/(2n+3)! solve using ratio test

Question

anonymous · Answer

attachment

irishboy123 · Answer

plodding mechanically through a definition i found on Wiki(!!!), ie

$$L = \lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|$$


 i get this...
 

$\lim\limits_{n \to \infty} \left| \dfrac{\dfrac{(-1)^{n+1} (n+1)!}{(2n+5)!}}{\dfrac{(-1)^n n!}{(2n+3)!}}\right|$

$= \lim\limits_{n \to \infty} \dfrac{(-1)^{n+1} ~ (n+1)! (2n+3)!}{(2n+5)! (-1)^n ~ n!}$

$ = \lim\limits_{n \to \infty} \left|  \dfrac{(-1) ~ n }{(2n+5) ~ (2n+4)} \right|$

$= \lim\limits_{n \to \infty}   \dfrac{  \frac{1}{n}}{(2+\frac{5}{n})(2+\frac{4}{n})} $

is that any help?!?!

if not, circle back :p

you should probs in future stuff things like this on the maths board, btw.  they usually have nothing better to do over there :-))