## anonymous 5 years ago Does this infinte series converge or diverge? Sigma from 1-infinity of 1n(n)/n^1.02

1. anonymous

n^(1.02)?

2. anonymous

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

3. anonymous

Use the integral test you will find it will converge.

4. anonymous

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

5. anonymous

In case you aren't If $\int\limits_{k}^{\infty} f(x) dx$ is convergent implies $\sum_{k}^{\infty}\an] is convergent 6. anonymous \[\sum_{k}^{\infty}an$ is convergent

7. anonymous

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

8. anonymous

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. =]

9. anonymous

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?

10. anonymous

Yes do it by parts

11. anonymous

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

12. anonymous

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

13. anonymous

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

14. anonymous

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

15. anonymous

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

16. anonymous

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

17. anonymous

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?

18. anonymous

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

19. anonymous

oh right, i see. thanks.

20. anonymous

Are we good?

21. anonymous

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

22. anonymous

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

23. anonymous

good luck!

24. anonymous

Thank You