Question

I can't figure out convergence and divergence tests, I'm going to fail my quiz if I don't figure this out. Please help someone.... Is the series 1/(ln2)^n convergent or divergent and if convergent, to what?

Answer 1

\[\sum_{1}^{\infty}\frac{ 1 }{ (\ln2)^n}\]

Answer 2

what is ln(2) ?

Answer 3

a number

Answer 4

roughly 0.69

Answer 5

yes, go on

Answer 6

what is 0.69 to a big power ?

Answer 7

a very large number?

Answer 8

so 1/ very large number is very small, to zero?

Answer 9

you can do this, if you don't have a calculator

0.7*0.7= 0.49 or about 0.5
0.7*0.7*0.7 is about 0.5*0.7 or 0.35
do you see a pattern ?

Answer 10

yes, when you multiply the a1 and a2 you get a3

Answer 11

is 0.69^3 bigger or smaller than 0.69^2

Answer 12

oops its smaller

Answer 13

I am not going to be able to go by that tho cuz we are not allowed calculators

Answer 14

and when n gets big you expect 0.69^n to get very small

Answer 15

sorry, I got disconnected. I still don't get what you are getting at

Answer 16

what type of series is it?
Is it geometric? If so, I can't see that in this a_n

Answer 17

to do this problem you need to know ln 2 is less than 1

a number between 0 and 1 will get smaller when you raise it to a power

Answer 18

does ln2 mean e^2?
I don't know what ln really means?

Answer 19

ln (short for natural log) means log base e


ln 2 is about 0.69 and e^0.69 is 2

but to finish this, 1/tiny number is a big number
the terms are getting larger as n gets bigger.

converge or diverge ? what do you think ?

Answer 20

** I should say e^(ln 2) = 2

Answer 21

you can also write \[\frac{1}{(\ln 2)^n}=\left(\frac{1}{\ln 2}\right)^n\]

Answer 22

the other way people check for convergence is using tests
the ratio test take the ratio of the n+1 term over the nth term
if you get a number bigger than 1 you know the sequence is diverging

Answer 23

this is insane, It will take me all night to type the equation. What does it converge to?

Answer 24

1/0.69 is about 1.44

You could estimate this by saying 1/.7 is 10/7 or 1 3/7

what is 1.44 raised to a big number?

Answer 25

But I have to show which test I use and what type of series it is. She never lets us convert ln to numbers.

Answer 26

hint: any number bigger than 1 gets bigger when you raise it to a power

if n is big enough, you will get huge numbers
as n-> infinity, 1.44^n also -> infinity

Answer 27

do I just plug in infinity? I thought we weren't supposed to have direct sub. when it makes the exponent infinity,

Answer 28

you can use the ratio test

however, you will have to know that ln 2 is less than 1 to know that this series diverges.

Answer 29

into the limit I meant

Answer 30

the -> means approaches

as n-> infinity, 1.44^n -> infinity

neither actually "get there"

but the point is, the terms are getting larger and larger, and the series diverges.

Answer 31

try this
\[\sum_{n=0}^\infty \frac{1}{(\ln 2)^n}=\sum_{n=0}^\infty \left(\frac{1}{\ln 2}\right)^n\] what kind or series is ^

Answer 32

damn thing won't show the equation you typed. I need a minute

Answer 33

I thought it weas a geometric series

Answer 34

right, it is geometric.

Answer 35

but it is not in the geometric form

Answer 36

what form do you mean ?

Answer 37

ar^n

Answer 38

a is 1, r is 1/ln(2) (roughly 1.44)

Answer 39

but the n is negative n

Answer 40

b^-n is the same as (1/b)^n

Answer 41

you are probably looking at (ln 2)^-n

you can also write it as (1/ln 2)^n

Answer 42

okay, so (1/ln)^-n is same as ln2^n??

Answer 43

\[ \frac{1}{(\ln 2)^n} = \left (\frac{1}{ \ln 2} \right)^n\]

Answer 44

as sirm already pointed out

Answer 45

oh, I see it now...

Answer 46

it is geometric, where a = 1 and r = (1/ln 2) as @phi gave

Answer 47

you definitely want to put this into a form that gives a positive n

Answer 48

ok I see this. Thank you

Answer 49

How can I give 2 medals??? grr

Answer 50

Can you do the ratio test on this series?

Answer 51

if I don't need to, isn't it harder?

Answer 52

First, it is good practice
Second, you said you have to show which test you used.