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.

See more answers at brainly.com

Join Brainly to access

this expert answer

SEE EXPERT ANSWER

To see the **expert** answer you'll need to create a **free** account at **Brainly**

go to the physics portion

what? no it's a math Q

but its in algebra 2

k sorry

Diverge means separate, converge means meet.

converge is to "settle down" on a finite falue

value*

oh jeez, an algebra2 definition of "diverge" ?
I kind of would like to know the context

Does it go to infinity?

Or does it go to a specific value?

or does it settle on no value at all and oscillate forever?

it goes infinitly

@nbouscal I wish they taught me that in algebra2, but I don't think so...

If it went to infinity then the sequence will increase exponentially

exactly^
@petegutz do you have a specific formula to use? there are a few...

no all that it gave me was what i put up

\[\sum_{n=1}^\infty ar^{n-1}=\sum_{n=0}^\infty ar^n=\frac a{1-r}\]is maybe a formula you can use?

identify |r| *

is it 1/5?

Yes, r=1/5

so it converges and it has a sum right?

Yes

Thanks!