# converges

• Dec 29th 2008, 03:19 PM
redieeee_babieeee
converges
ihi........ i have ths question where i have to decide whether it converges and if it does i need to state its limit.

an= 3n^4 -12 / 5n^3 + 8

i would devide it all by n giving 3n^3 - 12/n / 5n^2 + 8/n

but i do not know if this is the right approce andwhere to go after this.

thanx
• Dec 29th 2008, 03:22 PM
Jester
Quote:

Originally Posted by redieeee_babieeee
ihi........ i have ths question where i have to decide whether it converges and if it does i need to state its limit.

an= 3n^4 -12 / 5n^3 + 8

i would devide it all by n giving 3n^3 - 12/n / 5n^2 + 8/n

but i do not know if this is the right approce andwhere to go after this.

thanx

Is
$\displaystyle a_n = \frac{3n^4 -12 }{5n^3 + 8}$ ?

If so, then no.
• Dec 29th 2008, 03:23 PM
redieeee_babieeee
yes lol it is that
• Dec 29th 2008, 04:10 PM
Soroban
Hello, redieeee_babieeee!

Quote:

Decide whether it converges; if it does, state its limit.

. . $\displaystyle a_n\:=\: \frac{3n^4 -12}{5n^3 + 8}$

Divide top and bottom by the highest power of $\displaystyle n$ in the denominator, .$\displaystyle n^3$

$\displaystyle \lim_{n\to\infty}\frac{\dfrac{3n^4}{n^3} - \dfrac{12}{n^3}}{\dfrac{5n^3}{n^3} + \dfrac{8}{n^3}} \;\;=\;\;\lim_{n\to\infty}\frac{3n - \dfrac{12}{n^3}}{5 + \dfrac{8}{n^3}} \;\;=\;\;\frac{\infty - 0}{5 + 0} \:\;=\;\:\infty$ . . . diverges