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

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions.

- anonymous

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

- jamiebookeater

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- anonymous

n^(1.02)?

- anonymous

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

- anonymous

Use the integral test you will find it will converge.

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- 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. =]

- 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?

- anonymous

Yes do it by parts

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- 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?

- anonymous

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

- anonymous

oh right, i see. thanks.

- anonymous

Are we good?

- 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.

- anonymous

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

- anonymous

good luck!

- anonymous

Thank You

Looking for something else?

Not the answer you are looking for? Search for more explanations.