Question

determine whether the series is convergent or divergent

Answer 1

\[\sum_{n=2}^{\infty}\frac{ 6 }{ (n+2)(\ln(n+2)) }\]

Answer 2

@ganeshie8 does that mean that because "ln" is in this denominator, that it is automatically divergent?

Answer 3

It is divergent, yes. but having a "ln" in the denominator wont make it automatically divergent.

What does it mean for a series to diverge ?

Answer 4

1 + 1/2 + 1/3 + 1/4 + ...

does that series converge ?

Answer 5

1/n = diverge

Answer 6

^ that was the answer to your question above

Answer 7

thats right but that doesnt answer my qeuestion
im asking when do you say a series diverges ?

Answer 8

oh.. im not sure

Answer 9

Loosely speaking a `series` is "sum" of numbers

an `infinite series` is "sum" of infinite numbers

Answer 10

we say an `infinite series` converges when the "sum" converges to some number

otherwise we call it diverging

Answer 11

try the integral test, this was taylor made for that one (no pun intended)

Answer 12

i guess it was intended, since the expression should be "tailor made"

Answer 13

@ganeshie8 I see what you mean now

Answer 14

@satellite73 idk why but i suck at the integral test

Answer 15

make n in to x and integrate
this one is a straight up u -sub

Answer 16

hmm.. i have a lot to learn

Answer 17

\[\int_2^{\infty}\frac{dx}{(x+2)\ln(x+2)}\] put \(u=\ln(x+2)\) and you get the integral right away

Answer 18

\[6[( \ln (x+2) )( \ln (x+2)]\] evaluated from x-->2 ?

Answer 19

no, that is not the anti derivative

Answer 20

hi x!

Answer 21

YES, I am still here trying to understand how to do a problem like this

Answer 22

going by what satellite73 did
\[u=\ln(x+2) \Longrightarrow du=\frac{dx}{x+2}\]
\[x=2 \Longrightarrow u=2\ln 2\\ x=\infty \Longrightarrow u=\infty\]

Answer 23

hey what going on :)

Answer 24

just failing miserably over here

Answer 25

wouldnt a limit test be easier here? just asking

Answer 26

\[\int_{2\ln 2}^{\infty}\frac{du}{u}=\Large \left(\ln u \right]_{2\ln 2}^{\infty}\]

Answer 27

so we take the limit of lnu as u tends to oo

Answer 28

ooohhh.. you made that look way easier. I am not good with ln

Answer 29

apparently the limit is oo so this stuff diverge

Answer 30

wow....

Answer 31

thank u! it always helps me to SEE what you are thinking

Answer 32

that is what i was looking for

Answer 33

\[\lim_{u \to \infty}\ln u=\infty\]
you see it by graph