Ace school

with brainly

  • Get help from millions of students
  • Learn from experts with step-by-step explanations
  • Level-up by helping others

A community for students.

For what values of r>0 does the series converge?

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

Join Brainly to access

this expert answer

SIGN UP FOR FREE
I'm wary of using the integral test before actually getting some advices: |dw:1330831400158:dw|
yeah
.Sam. yeah what? Sorry to sound rude, but what's the point of disappointing me? I got so happy when I saw that notification pop up.

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

@Mertsj can u take a look?
uh I got it from a calculator it says that its converge but I don't know how it solves, lol
@malevolence19 can u help?
sry for the disappointment :D
tnx for tryin sam
Okay, I think I can help you.
awesome
For convergence of a series using the ratio test we must have that: \[\lim_{n \rightarrow \infty}a_n \rightarrow 0\] And: \[\lim_{n \rightarrow \infty} \left| \frac{a_{n+1}}{a_n}\right|<1\] So: \[\lim_{n \rightarrow \infty} \left| \frac{r^{\ln(n+1)}}{r^{\ln(n)}}\right|=\lim_{n \rightarrow \infty} \left| r^{\ln(n+1)-\ln(n)}\right|=\lim_{n \rightarrow \infty} \left| r^{\ln \left( \frac{n+1}{n}\right)}\right| \rightarrow 1\] Means that the ratio test is inconclusive. We need to find another approach (I should have realized this wouldn't work, its a rational function :/)
@JamesJ help?!
mal19 almost had it. If |r| < 1, then the limit of the ratio a_{n+1}/a_n converges to zero.
oh so that's it?
It's that simple? dang it and i'm sitting here with improper integrals!
*correction: If |r| < 1, then the limit of the ratio a_{n+1}/a_n converges to a number less than 1. And then by that ratio test, the sum converges.
U guys rule!
namely, it converges to |r| itself.

Not the answer you are looking for?

Search for more explanations.

Ask your own question