Question

Determine whether the following sequences converge or diverge.

Answer 1

7(-1)^(6n)

Answer 2

Since n will always be a multiple of 6 (and 2)... then this sequence will be converging to 7

Answer 3

2) an = (5/8)^n

Answer 4

I think it diverges, however I do not know how to explain this one.

Answer 5

3) an = 3^n / 5^(n+1); converges to 0? denominator is always greater than numerator?

Answer 6

^ would we have to use LHospital's Rule

Answer 7

\[a_n=7(-1)^{6n} \\ a_n=\left\{ 7,7,7,7,... \right\}\]
I would definitely say the first sequence converges since the numbers are constantly 7 as you say the exponent is a multiply of 2 always.

Answer 8

^ Awesome :)

Answer 9

Have you thought about the next one ? :)

Answer 10

The second one did you use the geometry series thingy

Answer 11

geometric*

Answer 12

Not sure what that is...

Answer 13

oops sorry that isn't a series
lol it is a sequence nevermine

Answer 14

nevermind

Answer 15

I mean I know what it is, but I didn't know how to apply it. Oh okay

Answer 16

\[a_n=(r)^n \\ \]
an convereges if r is between -1 and 1 (including 1)
and diverges for other value of r

Answer 17

r=5/8 in your second sequence

Answer 18

so what do you think again about the second sequence?

Answer 19

I said it diverged like you said!

Answer 20

Now I am not sure about the 3rd one...

Answer 21

oh no i didn't say it divereges

Answer 22

5/8 is between -1 and 1

Answer 23

so (5/8)^n convereges

Answer 24

Ohhh! I was thinking of the equation being rewritten as 5^n / 8^n

Answer 25

Would that work also?

Answer 26

\[(\frac{5}{8})^n=\left\{ \frac{5}{8},\frac{25}{64},\frac{125}{512},... \right\}\]
these numbers are getting closer to 0 aren't they?

Answer 27

Ahhaahaha Okay !

Answer 28

\[\approx \left\{ 0.625, 0.390625, 0.244140625, 0.152587890625, 0.095367431650625,...\right\}\]
I wrote out the first few approximations

Answer 29

so it might be clearer

Answer 30

Thank you :) What about the third sequence? 2^n / n+1?

Answer 31

3) an = 3^n / 5^(n+1); converges to 0? denominator is always greater than numerator?

this one or that one?

Answer 32

2^n / 5^(n+1)

Answer 33

yess

Answer 34

\[a_n=\frac{3^n}{5^{n+1}} =\frac{3^n}{5^n 5^1}=\frac{1}{5} \cdot \frac{3^n}{5^n}=\frac{1}{5}(\frac{3}{5})^n\]

Answer 35

you can use that theorem I mentioned earlier

Answer 36

a_n convereges if r is between -1 and 1 (including 1)

Answer 37

a_n diverges if r is not in that interval

Answer 38

so r=3/5
that means (3/5)^n does what?

Answer 39

Converges! Thank you

Answer 40

yes and 1/5 is just a constant multiple
so if (3/5)^n converges then 1/5*(3/5)^n also converges