Question

Someone check this calc 2?

Answer 1

I said converges since I got a limit at 0?

Answer 2

Correct

Answer 3

how did you get 0?

Answer 4

\[\lim_{n \rightarrow \infty}n(n-6)\]
that looks like a product of really big numbers

Answer 5

Did i mess up? let me try again...

Answer 6

@Haseeb96 @freckles so am i right or not guess we can check on wolfram

Answer 7

the product of really big positive numbers is a really big positive number

Answer 8

not 0

Answer 9

So the limit is infinity? I mean, as long as there is a limit it converges, right?

Answer 10

infinity or -infinity means it diverges

Answer 11

if the limit is infinity then it will be divergence

Answer 12

infinity isn't a number just so you know
it just means it gets really really big

Answer 13

but @freckles he said limit is 0
so i said he is correct

Answer 14

why is the limit 0 @Haseeb96

Answer 15

Ok. Well, thanks all. I need to go back and review I guess.

Answer 16

I'm pretty sure the product of really big positive numbers can not be 0

Answer 17

do you understand why it diverges @AmTran_Bus

Answer 18

Yes, I understand if it is infinity it diverges. Thanks. But I need to solve the limit correctly.

Answer 19

But I see what you are saying with that @freckles

Answer 20

\(a_n=n(n-6)\)

\(\displaystyle \lim_{n\rightarrow\infty }n(n-6)\)

\(\displaystyle \lim_{n\rightarrow\infty }n^2-6n\)

diverges to positive inifinity.

Answer 21

Thanks solomon

Answer 22

Sure.... everytime to see the sequence convergence, for any sequence \(A_n\) , take the limit of it as n approaches infinity

Answer 23

60(54)=?
100(94)=?
1000(994)=?
10000(9994)=?
these products are getting super super big

Answer 24

And if sequence diverges, then (always!) the series diverges

Answer 25

Yes. I super see that now. Thanks so much freckles. I think I understand it.

Answer 26

Thanks solomonzelman

Answer 27

yw

Answer 28

((that was my favorite section when I learned it, btw))

Answer 29

good luck, you will get an A I am sure... !