Question

Determine the convergence or divergence of the series

Answer 1

break the log in two parts

Answer 2

then it should be telescopic

Answer 3

maybe

Answer 4

chain rule is a derivative/integral thing

Answer 5

I was thinking it was telescopic

Answer 6

log(a/b) = log(a) - log(b)

Answer 7

So i would have ln(1)-ln(n)

Answer 8

log(1) - log(n) + log(1) - log(n+1) + log(1) - log(n+2)

log(1) is zero, so this is the sum of -log(n)

Answer 9

we could have converted it to -log(n) from the get go if i had realized log(n^-1)

Answer 10

any ideas what this is leadinng to?

Answer 11

-log(n)-log(n+1)-log(n+2).....

Answer 12

yeah, i meant with regards to converge or diverge

Answer 13

does the sequence generated by log(n) tend towards 0 as n gets bigger? or does it grow without bounds

Answer 14

Bigger with out bounds

Answer 15

then adding up all those bigger and bigger values, this thing isnt going to converge

Answer 16

can we use squeeze theorem??

Answer 17

Thank you

Answer 18

the limit test, the wolf says, shows divergence

Answer 19

Since ln (1/n) < 1/n for all n >0

Answer 20

I am so lost on how the squeeze theorem

Answer 21

works

Answer 22

and claim 0< ln(1/n)< 1/n
Let take lim 1/n when n goes to infinitive =0, and lim 0 =0 so that lim ln(1/n) =0

Answer 23

ln(1/n) < 0 for large n>1

Answer 24

If the limit = 0 then per the divergence test the series converges.

Answer 25

\[0< ln \dfrac{1}{n}<\dfrac{1}{n}\]
lim 0 =0
lim 1/n =0
so the lim of middle term =0

Answer 26

again, ln(1/n) < 0

Answer 27

ln(1/2) < 0

ln(1) = 0

ln(2) > 0

Answer 28

oh, yes, I am sorry.

Answer 29

other than that :) it was good

Answer 30

Thanks for the help