Question

Does this infinte series converge or diverge? Sigma from 1-infinity of 1n(n)/n^1.02

Answer 1

n^(1.02)?

Answer 2

\[\sum_{1}^{\infty} \ln(n)/n^1.02\]

Answer 3

Use the integral test you will find it will converge.

Answer 4

Are you familiar with integral test? Or should I go on?

Answer 5

In case you aren't
If \[\int\limits_{k}^{\infty} f(x) dx \] is convergent implies \[\sum_{k}^{\infty}\an] is convergent

Answer 6

\[\sum_{k}^{\infty}an\] is convergent

Answer 7

take \[f(x) = \frac{\ln(n)n^(1.02)}\] and k = 1

Answer 8

f(x) = ln(n)/n^(1.02) take the integral from 1 to inf and show it does not diverge that is it has a finite answer. That will mean our series also converges. =]

Answer 9

i am also interested in this problem, and i'd love to use the integral test, but i don't know how to integrate that. would you do it by parts?

Answer 10

Yes do it by parts

Answer 11

let u=ln(n), dv=n^(1.02)? see but from here what would v be?

Answer 12

sorry, i don't mean to hijack the problem, it's just the same thing i've been having trouble with, too.

Answer 13

oh and i meant dv=n^(-1.02)dx

Answer 14

unless i'm setting u and dv to the wrong parts

Answer 15

v is something stupid -50/n^(0.02)

Answer 16

...what. where did the 50 come in?

Answer 17

i can see the power rule making the exponent come down to 0.02, and also kicking the negative out front, but 50? where did that come from?

Answer 18

1.02 = 1 + 1/50, sorry I hate decimals

Answer 19

oh right, i see. thanks.

Answer 20

Are we good?

Answer 21

oh i am so good with this now. but then i definitely stole this question from the original poster, so... sorry about that, Aubun.

Answer 22

Okay! I'm off to bed, have a big math test tomorrow

Answer 23

good luck!

Answer 24

Thank You