# Calc 2 Sequence

• Nov 15th 2007, 01:33 PM
poopforbrains
Calc 2 Sequence
Can anyone lend me a hand

Write the first 5 terms of the sequence

an = 3n!/(n-1)!

Find the limit of the sequence if the limit exists

an = (n-2)!/n!
• Nov 15th 2007, 04:56 PM
ThePerfectHacker
Quote:

Originally Posted by poopforbrains
Find the limit of the sequence if the limit exists

an = (n-2)!/n!

$\displaystyle \frac{(n-2)!}{n!} = \frac{(n-2)!}{n(n-1)(n-2)!} = \frac{1}{n(n-1)}$.
• Nov 15th 2007, 05:08 PM
Soroban
Hello, poopforbrains!

Quote:

Write the first 5 terms of the sequence: .$\displaystyle a_n \:= \:\frac{3n!}{(n-1)!}$

You're in Calculus 2 and you can't plug in values ??

. . $\displaystyle a_1 \: = \: \frac{3\cdot\!1!}{0!} \: = \: 3\cdot1 \: = \: 3$

. . $\displaystyle a_2 \: = \: \frac{3\!\cdot\!2!}{1!} \: = \: 3\cdot2 \: = \: 6$

. . $\displaystyle a_3 \: = \: \frac{3\!\cdot\!3!}{2!} \: = \: 3\cdot3 \: = \: 9$

. . $\displaystyle a_4 \: = \: \frac{3\!\cdot\!4!}{3!} \: = \: 3\cdot4 \: = \: 12$

. . $\displaystyle a_5 \: = \: \frac{3\!\cdot5!}{4!} \: = \: 3\cdot5 \: = \: 15$

Quote:

Find the limit of the sequence if the limit exists
. . $\displaystyle a_n \:= \:\frac{(n-2)!}{n!}$

We have: .$\displaystyle a_n \;=\;\frac{(n-2)!}{n(n-1)(n-2)!} \;=\;\frac{1}{n(n-1)}$

Hence: .$\displaystyle \lim_{n\to\infty}\,\frac{1}{n(n-1)} \;=\;0$