Calc 2 Sequence

**poopforbrains** · November 15th 2007, 01:33 PM

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!

**ThePerfectHacker** · November 15th 2007, 04:56 PM

Originally Posted by poopforbrains

Find the limit of the sequence if the limit exists

an = (n-2)!/n!

$\frac{(n-2)!}{n!} = \frac{(n-2)!}{n(n-1)(n-2)!} = \frac{1}{n(n-1)}$ .

**Soroban** · November 15th 2007, 05:08 PM

Hello, poopforbrains!

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

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

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

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

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

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

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

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

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

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

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