Question

find the sum of the series if it converses or show it diverges.

Answer 1

\[\sum_{n=0}^{infinity} (4^{3n}-40)/3^{4n}\]

Answer 2

have you tried the ratio test? or any of the other handful of tests?

Answer 3

ive no idea how to fine the limit 1st...

Answer 4

lets split the fraction into 2 parts, and see if they work out individually

Answer 5

\[\frac{4^{3n}}{3^{4n}}-\frac{40}{3^{4n}}\]

Answer 6

yea the 2nd part =0 but the 1st part.....

Answer 7

the first part, try flipping it, subtracting 1 from the n parts, and multplying them together to take a limit

Answer 8

\[\frac{4^{3n}}{3^{4n}}\frac{3^{4(n-1)}}{4^{3(n-1)}}=\frac{4^{3n}}{3^{4n}}\frac{3^{4n}3^{-1}}{4^{3n}4^{-1}}\]
\[\frac{\cancel{4^{3n}}}{\cancel{3^{4n}}}\frac{\cancel{3^{4n}}3^{-1}}{\cancel{4^{3n}}4^{-1}}=\frac{4}{3}\]

Answer 9

id have to read up on the rules, but i think that means it diverges if it doesnt go to zero

Answer 10

http://www.wolframalpha.com/input/?i=sum%28from+0+to+inf%29+%284%5E%283n%29-40%29%2F3%5E%284n%29

err, converges :) i guess if its to inf it diverges, to 0 inconclusive; and anything else convergent

Answer 11

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

this would be a better read up of the rules instead of me trying to remember them incorrectly

Answer 12

ok, im still trying to figure it out how u flip it.~~

Answer 13

this might help; you are trying to determin how the end of it behaves - way off near the infinities.

the ratio of the now term with respect to the next term is looked at:

\[\frac{now}{next}\]

since you have a fractional representation of the series it follows that:

\[\frac{now/a}{next/a}=\frac{now}{a}\frac{a}{next}\]

Answer 14

well, next/now that is

Answer 15

allow me to butt in

Answer 16

allowance granted lol

Answer 17

allow me to shut up and read more carefully

Answer 18

in my train of reasoning: if we look at descending values:

10,8,6,4,2 and want to know if they are getting smaller; the ratio of next/now is what we are comparing

8/10 6/8 4/6 2/4
=4/5 =3/4 =2/3 =1/2

the ratios are getting smaller and smaller

Answer 19

take the sequence: 1,3,5,7,9,11

if the terms are getting larger, or ascending. we get values that are getting bigger and bigger, thier next/now ratio gets larger as well

3/1 5/3 7/5 9/7 and will always be larger than 1

Answer 20

wow, im getting to it.~~ looks good

Answer 21

youre setup is messier than these pretty integers; but the same concept applies.

divide the next by the now and take the limit of that ratio as the ns approach infinity; if taht value is greater then 1, they are getting bigger and diverge

if it is less than 1, they are getting smaller and converge

if they are equal to 1, the test is inconclusive and something else would need to be tried

Answer 22

allow me to butt in again

Answer 23

please do, my memory tends to be like swiss cheese these days :)

Answer 24

\(4^{3n}=64^n\)
\[3^{4n}=81^n\]

Answer 25

converges because it is a geometric series with \(r=\frac{64}{81}<1\)

Answer 26

that is a viable test yes

Answer 27

the clue was "add" and since there are very few tools at our disposal to add these (can't turn it in to partial fractions and get collapsing sum for example) figured the gimmick was to make the bases the same in order to add a geometric series

Answer 28

unfortunately it is too early for me to figure out what to do with that annoying \(-40\)

Answer 29

http://www.wolframalpha.com/input/?i=%284%5E%283%28n%2B1%29%29%2F3%5E%284%28n%2B1%29%29%29%2F%284%5E%283n%29%2F3%5E%284n%29%29

since i cant do math, ratio according to wolf workks out to 64/81 as well :)

and the -40/inf goes to 0 so this converges

Answer 30

3(n-1) = 3n-3 NOT 3n-1 ;) ugh

Answer 31

ok finally figured out how to add this

Answer 32

\[\sum(\frac{64}{81})^n-40\sum(\frac{1}{81})^n\] two geometric series does it