Question

group theory: test

Answer 1

@TonyJOu

Answer 2

ok tony stay in here

Answer 3

ok

Answer 4

\[\sum_{n=1}{\frac{1+2^{?}}{3^{n}}}\]

Answer 5

I cant see on my phone

Answer 6

that's almost there. It goes to infinity and the A sub n is 1 + 2^(n-1) / 3^n

Answer 7

ok, what

Answer 8

ok, what all test can you use?

Answer 9

I was thinking you could use Test for Divergence but I was thrown off because the numerator was 1+2^(n-1). Had it just been 1+2^n, I would have a limit of 2/3 because the degrees were the same.

Answer 10

\[\frac{1+2^{n-1}}{3^n}<\frac{2+2^{n-1}}{3^n}=\frac{2^1+2^{n-1}}{3^n}=\frac{2^{n}}{3^n}\]

Answer 11

and that is a geometric that converges, thus it converges by the comparison test.

Answer 12

lol, you are typing then erasing allot:)

Answer 13

it's the way that i'm trying to type this crap in. ;)

Answer 14

Where did the 2 come from?

Answer 15

I know, just type it as you would a ti-89

Answer 16

well I just used that to show that your sequence is less than (2/3)^n

Answer 17

The initial problem was 1+2^(n-1), then I noticed a 2 all of a sudden.

Answer 18

it didn't come from anywhere, its just the fact that
\[\frac{1+2^{n-1}}{3^n}<\frac{2+2^{n-1}}{3^n}\]

Answer 19

So working with the geometric rules, are you stating that something converges if it is greater than or equal to -1 and less than or equal to 1?

Answer 20

we used algebra to show that your sequence a_n < (2/3)^n
so sum a_n < sum (2/3)^n
since we know sum (2/3)^n converges, we know that by the comparison test sum a_n must converge

Answer 21

so it converges to 2/3??

Answer 22

you are supposed to find the sum?

Answer 23

Yeah it said if it it's convergent, determine the sum.

Answer 24

ahh
well
(1+2^(n-1))/3^n = (1/3)^n+(2^(n-1))/3^n = (1/3)^n + (1/2)(2/3)^n

Answer 25

let me write this up and take a picture

Answer 26

ok

Answer 27

let me know if this makes since

Answer 28

give me a few minutes to mull it over. I didn't realize it was so involved.

Answer 29

its not really, its just break apart the fraction and then lil manipulation to get (2/3)^n

Answer 30

yeah take your time ill be around for a few hours

Answer 31

where did the ^n-1 that was in the numerator? In the third line down, I noticed it was gone. It became ^n.

Answer 32

well we have \[\frac{2}{2}*\frac{2^{n-1}}{3^n}=\frac{1}{2}\frac{2^1*2^{n-1}}{3^n}=\frac{1}{2}\frac{2^{n-1+1}}{3^n}\]

Answer 33

ahhhh I got ya. I see.

Answer 34

I create the fraction (2/2) which is just 1, so it does not hurt to multiply by it
then I factor out the bottom 2, and use the top one

Answer 35

you will use that trick allot..

Answer 36

And on the 4th line down, shouldn't it be (1/3)/1-(1/3)?

Answer 37

there was need to use the first method I showed with the comparison test, but make sure you understand it. this is a common thing to do when you need to show something is less than something else. (specifically that part where the 1 becomes 2).

Answer 38

it is, it just looks like 2

Answer 39

\[\frac{\frac{1}{3}}{1-\frac{1}{3}}+\frac{1}{2}\frac{\frac{2}{3}}{1-\frac{2}{3}}\]

Answer 40

to do a fraction
\[\frac{1}{2}\.]

Answer 41

without the .

Answer 42

anyway, ok so you got it?

Answer 43

yeah man I got it. This is so much harder than 252 and 251. I understand this, it's just taking me a while with all these different tests for convergence and divergence.

Answer 44

yeah. well it goes to show that there are many different ways to show something converges. I first showed it with the comparison test, and then we used the fact that it was the sum of two geometric series. But once you told me that you needed to find the sum, I know then it must be geometric or telescopic (because these are the only two we know how to find the sum of). we know its not telescopic because its always positive. thus it must be geometric.

Answer 45

What is the test of two geometric series then? Meaning that once you knew we had to find the sum, where is the test for that?