Question

can someone help me

Answer 1

@jhonyy9

Answer 2

@AZ

Answer 3

@dude

Answer 4

@Angle

Answer 5

so it's been years since I've done this but until someone more competent can assist, I shall try my best with whatever I remember/ have just reviewed right now while using the assistance of today's technology (:

We first need to determine if the series converges or diverges.
If it converges, then we need to determine if it absolutely converges or conditionally converges.

For the first question, does it converge or diverge?

Try using the integral test.

Answer 6

So does this integral diverge or converge?

\[\int\limits_{2}^{\infty}\left(\frac{1}{x\log\left(x\right)\log\left(\log\left(x\right)\right)}\right)dx\]

Answer 7

going to tag other users who might remember this material much better than me to take a look/take over whenever (or if) they're available
@Hero @imqwerty @jimthompson5910 @Vocaloid

Answer 8

i was able to do part a already

Answer 9

mhmm well I guess for part A, you could have just done the comparison test like it said so since

\(\sum\limits_{n = 2}^\infty {\frac{1}{{n\ln n}}}\) is divergent, then your series in (a) is also divergent

but anyway, which part are you doing right now?

Answer 10

try for k=1 what you get ?

Answer 11

wouldnt we plugin k = 0?

Answer 12

\(\color{#0cbb34}{\text{Originally Posted by}}\) @xXMarcelieXx
im stuck on c
\(\color{#0cbb34}{\text{End of Quote}}\)
have you tried checking for absolute convergence?

Answer 13

and write what you get for k=2

assume the first 2 terms

try for k=3 what you get add to the sum of first 2 terms and see the sum go to a fixed value or not
(\color{#0cbb34}{\text{Originally Posted by}}\) @xXMarcelieXx
wouldnt we plugin k = 0?
\(\color{#0cbb34}{\text{End of Quote}}\)
yes the first term will be for k=0

Answer 14

\(\color{#0cbb34}{\text{Originally Posted by}}\) @AZ
So does this integral diverge or converge?

\[\int\limits_{2}^{\infty}\left(\frac{1}{x\log\left(x\right)\log\left(\log\left(x\right)\right)}\right)dx\]
\(\color{#0cbb34}{\text{End of Quote}}\)
++
this integrates to \(log(log(log(x)))\) which approaches infinity as x -> infinity.

Answer 15

for k = 0 i got 1 and k = 1 we get -1/2

Answer 16

for k=2 ?

Answer 17

k-=2 is 1/4

Answer 18

ok we start with

1 the first term
1+(-1/2)=1/2 the sum of first two terms
the 3rd term is 1/4
1/2 +1/4 = 3/4 the sum of first 3 terms

do you see where go the sum ? and from this what you think divergent or convergent ?

Answer 19

convergent?

Answer 20

so what is the difference between a convergent and a divergent line ?

Answer 21

convergent means that it approaches to a number and divergent approaches close to infinity

Answer 22

yes exactly so this sum where you see that go - approaches - ?

Answer 23

to a fix number or infinity ?

Answer 24

\(\color{#0cbb34}{\text{Originally Posted by}}\) @jhonyy9
ok we start with

1 the first term
1+(-1/2)=1/2 the sum of first two terms
the 3rd term is 1/4
1/2 +1/4 = 3/4 the sum of first 3 terms

do you see where go the sum ? and from this what you think divergent or convergent ?
\(\color{#0cbb34}{\text{End of Quote}}\)
look here pls

Answer 25

1/2
3/4

are fix numbers ?

Answer 26

hmm yes

Answer 27

so then ?

Answer 28

\(\color{#0cbb34}{\text{Originally Posted by}}\) @xXMarcelieXx
convergent?
\(\color{#0cbb34}{\text{End of Quote}}\)
you answered it here

Answer 29

but you not was sure in this

Answer 30

it would be convergent right?

Answer 31

exactly

Answer 32

bc has a fix value

Answer 33

bc this sum is a fix number

Answer 34

do you understand it now ?

Answer 35

yes, but how can i show this in proof form?

Answer 36

how i ve wrote above

first term is 1
second is -1/2 so the sum will be 1 + (-1/2) = 1/2
3rd term is 1/4 so the sum of first 3 terms will be
1/2 + 1/4 = 3/4
do you need - wan - continue ?
then you need calcule the 4th term
the 5th term
the 6th term
and so ...
but dont forget to assume them

Answer 37

what do you mean by assume them

Answer 38

add them bc. you need calcule the sum of them

Answer 39

@AZ any idea pls ?

Answer 40

For c, Use the root test on the absolute series

Answer 41

A series \(\sum a_n\) is absolutely convergent if \(\sum \left|a_n\right|\) is convergent.