Question

Help..

Answer 1

How can i apply integral test to this

\[\frac{ 1 }{ k( Logk)^{p}}\]

Answer 2

what are the limits ?
2 to infinity ?

Answer 3

yes..

Answer 4

\(\large \sum \limits_2^{\infty } \frac{1}{k (\ln k)^p} \) converges \(\iff\) \(\large \int \limits_2^{\infty} \frac{1}{k(\ln k )^p}dk \) is finite

Answer 5

\(\large \sum \limits_2^{\infty } \frac{1}{k (\ln k)^p} \) diverges \(\iff\) \(\large \int \limits_2^{\infty} \frac{1}{k(\ln k )^p}dk \) is not finite

Answer 6

For which p the integral is finite ?

Answer 7

evaluate the right side integral and see when it is finite

Answer 8

p is greater than 1 ?

Answer 9

ln k = u
1/k dk = du

\(\large \int \limits_{\#}^{*} 1/t^p dt\)

analyze this

Answer 10

\(\large \frac{t^{-p+1}}{-p+1} \Big|_{\ln 2}^{\infty } \)

Answer 11

when p < 1, clearly it diverges

Answer 12

p =1 also it diverges

p > 1 is when it wud be finite,

Answer 13

let me know if u want discuss the reasoning behind these 3 conclusions

Answer 14

why at p=1 diverges? because its result becomes 0/0 then the integral is undefined ?

Answer 15

nope

Answer 16

\(\large \int \limits_{\ln 2}^{\infty} 1/t dt = \ln t \Big|_{\ln 2}^{\infty} = \ln \infty - **** = \infty \)

Answer 17

so, p = 1 also diverges

Answer 18

thank you..