Question

Determine whether the sequence converges or diverges. If it converges, give the limit.

60, -10, 5/3, -5/18 ...

Answer 1

Looks like \(\left\{a_n\right\}=\left\{(-1)^{n+1}\dfrac{60}{6^{n-1}}\right\}_{n=1}^\infty\).

Find \(\displaystyle \lim_{n\to\infty}a_n\).

Answer 2

what? @SithsAndGiggles I'm a bit confused

Answer 3

The first thing I did was write the nth term. Another way of expressing the same sequence.

You have alternating signs of successive terms, so you know there's a power of (-1) somewhere along the line, such as \((-1)^n\) or \((-1)^{n+1}\). Which of these you use depends on the sign of the first term.

Your first term is positive 60. Starting your sequence with n=1, the only way to start with a positive number is if you have an even power of -1, so you use \((-1)^{n+1}\). (Plugging in n=1 gives you \((-1)^2=1\), thus a positive sign.)

Now let's ignore the sign for now, since it's been taken care of. The first term is 60, then 60/6, then 60/36, and so on. In the denominator you have a power of 6 (\(6^n\)); each successive term is the previous term divided by 6.

You start with \(\dfrac{60}{1}=\dfrac{60}{6^0}\). Since we're starting the sequence with n=1, we change the exponent on the 6 to n-1 to include this first term.

So you're left with
\[a_n=(-1)^{n+1}\frac{60}{6^{n-1}}\]
Does that make sense?

Answer 4

okay yes that makes sense! from that point how do I find the limit and decide if it converges or diverges

Answer 5

@SithsAndGiggles

Answer 6

\[\lim_{n\to\infty}a_n=\lim_{n\to\infty}(-1)^{n+1}\frac{60}{6^{n-1}}\]
Do you need help with the actual limit-finding?

Answer 7

let me try it real quick

Answer 8

oh wait yeah I need help finding it.. I'm not sure what to do with n

Answer 9

\[\lim_{n\to\infty}(-1)^{n+1}\frac{60}{6^{n-1}}\]
The \((-1)^{n+1}\) part will alternate between -1 and 1 as n becomes arbitrarily large, so think of the limit as
\[\lim_{n\to\infty}\pm\frac{60}{6^{n-1}}\]
The bottom will approach ∞, so...

Answer 10

I am so lost!

Answer 11

:( blehhh I'm going to die

Answer 12

Back up a bit. According to the directions, you have to show whether the sequence is convergent, then provide the limit. Do you know how to show it's convergent?

Answer 13

No I have no idea, I'm doing this class on my own and nothing covered it in my text it just went over geometric sequences :(

Answer 14

That's actually exactly what you need! This sequence is geometric. What do you know about the convergence of a geo seq?

Answer 15

I have to brb for about 15 minutes will u be on when i get back? I'm learning a lot from your explanation!

Answer 16

Sure, I can wait.

Answer 17

While I am waiting, I suppose I can write some more...
\[\left\{a_n\right\}=\left\{(-1)^{n+1}\dfrac{60}{6^{n-1}}\right\}_{n=1}^\infty\]
See why it's a geo seq?
\[(-1)^{n+1}\dfrac{60}{6^{n-1}}=(-1)^{n+1}\frac{60\cdot6}{6^n}=360(-1)^{n+1}\left(\color{red}{\frac{1}{6}}\right)^n\]
So you have a geo seq with \(r=\dfrac{1}{6}\). Does it converge or diverge? Once you answer this question, go back to finding the limit.