Question

Determine and show if this series is convergent or divergent

an = 4^(n+3)/3^(n+4)
so I have

4^(3) (lim 4^(n)) / 3^(4)(lim 3^(n))
n->infinity n->infinity

How would I compute the limit?

Answer 1

it seems to me that it would be 4^(3)/3^(4) but wolfram alpha is giving me zero as the limit anyone able to explain to me why that is?

Answer 2

wait it displays monotonicity

(4^(3)/3^(4)) lim (4/3)^(n)
n-> infinity

Answer 3

I cant use squeeze law on it though

Answer 4

because there is no upper bound

Answer 5

Find \(\Large \frac{a_{n+1}}{a_n}\), if it's a constant, then the series will converge.
Could you check it?

Answer 6

(4^(n+1)/3^(n+1))/4^(n)/3^(n)
=
4^(n+1)3^(n)/3^(n+1)(4^(n))
=
4^(n)(4)(3^(n))/3^(n)(3)(4^(n))
=
4/3

Answer 7

what rule is that?

Answer 8

I can see the function is increasing but I dont think it has a lower bound or upper bound

Answer 9

its the ration test\[\]

Answer 10

ratio ....

Answer 11

I also have to compute the limit

Answer 12

and using a convergence method that i have in my head (could be wrong tho)\[\lim_{n\to\ inf}\frac{4^{n+3}}{3^{n+4}}=\lim_{n\to\ inf}\frac{4^{n}4^{3}}{3^{n}3^{4}}\]

\[K\ \lim_{n\to\ inf}(\frac{4}{3})^n\]

Answer 13

right thats what I have so far but I don't know where to go from there

Answer 14

also for that test to work doesn't it need to have an upper or lower bound?

Answer 15

It is pretty clear that the limit goes to 0 but how do I show that

Answer 16

Lhopital perhaps?

Answer 17

or integral rule?

Answer 18

do I use the rule

Kln(L) = nln(3/4)

Answer 19

sorry I mean

Kln(L) = K(nln(3/4))

Answer 20

then convert it into a fraction and use L'H

Answer 21

dunno, that looks foreign to me

4^n/3^n is already a fraction; and I dont really see it Lhoping out

Answer 22

true, ok I will try that, well it is a rule apparently I have it in my notes

Answer 23

derivatives will always end up with (4/3)^n

Answer 24

any exponential function with a base greater then 1 diverges i think

Answer 25

I think I need to use that rule otherwise I get 4^(n)log(4)/3^(n)log(3)

Answer 26

http://www.wolframalpha.com/input/?i=limit+4%5E%28x%2B3%29%2F3%5E%28x%2B4%29

perhaps, but you might wanna chk this out to chk things by

Answer 27

and it just will get more complex from there

Answer 28

weird it told me the limit was 0 when I put it into wolfram alpha

Answer 29

n to -inf = 0

n to +inf = inf

n to 0 = ....what it said

Answer 30

so it does go to infinity

Answer 31

so it is divergent

Answer 32

correct, it diverges.

Answer 33

what is the test I can apply?

Answer 34

ratio test should do it

Answer 35

doesnt it need an upper or lower bound though?

Answer 36

thats the one ash presented at the start

Answer 37

no

Answer 38

ratio test looks at how fast the slope is changeing at the ends

Answer 39

so if an+1/an > 1 then it is divergent?

Answer 40

lets see :)

4^(x+3) * 3^(x+3)
----------------- = 4/3
4^(x+2) * 3^(x+4)

i think theres a different "rule"; but so far its seems right

Answer 41

http://tutorial.math.lamar.edu/Classes/CalcII/RatioTest.aspx

this should be more definitive than I can recall

Answer 42

c < 1 < d

if 1, then try different test

Answer 43

thanks

Answer 44

yw