Question

help!!

Answer 1

For a series to be convergent, \(\Large\displaystyle\lim_{n\rightarrow\infty}\left|\frac{a_{n+1}}{a_n}\right|<1\)

Answer 2

That means \(\Large\displaystyle\lim_{n\rightarrow\infty}\left|\frac{\frac1{(n+1)(\ln (n+1))^p}}{\frac1{n(\ln n)^p}}\right|<1\)

Answer 3

how do u find p from ratio test? i got n/(n+1)(ln n)^p

Answer 4

Where did the ln(n+1) go? :O

Answer 5

i thought that would be canclled out by ln n haha

Answer 6

okay i might be wrong for that. pls, show me how T^T

Answer 7

Well you'd get n/(n+1)[ln(n)/ln(n+1)]^p

Answer 8

okay, then after that? there are 2 variables, n and p

Answer 9

n is set to infinity

Answer 10

I think l'hopital should be used here?

Answer 11

that would be a lttle complicated

Answer 12

Well...

Answer 13

You do notice n/(n+1) = 1

Answer 14

yes..

Answer 15

You do notice (ln n)^p is differentiable

Answer 16

u mean (lnn/ln(n+1))^p

Answer 17

No coz l'hopital

Answer 18

i'm confused, can you break it down clearly/

Answer 19

Well I'm sick so my brain ain't propering functionly

Answer 20

\[\lim_{n \rightarrow \infty} \frac{ n }{ (n+1)[\frac{ \ln n }{ \ln (n+1) }]^p }\]

Answer 21

hmm okay then,

Answer 22

n/(n+1)[(ln n)^p/(ln n+1)^p]

Answer 23

i.e. [(ln n)^p/(ln n+1)^p]

Answer 24

l'hopital: [p(ln n)^(p-1)/n]/[p(ln n+1)^(p-1)/n]

Answer 25

so it will be pln^(p-1)/pln(n+1)^p-1

Answer 26

all that <1 ?

Answer 27

Well

Answer 28

I actualy don't know... I'm so dizzy now

Answer 29

i'm sorry u shd go rest