Someone check this calc 2?

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Someone check this calc 2?

Mathematics
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I said converges since I got a limit at 0?
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how did you get 0?
\[\lim_{n \rightarrow \infty}n(n-6)\] that looks like a product of really big numbers
Did i mess up? let me try again...
@Haseeb96 @freckles so am i right or not guess we can check on wolfram
the product of really big positive numbers is a really big positive number
not 0
So the limit is infinity? I mean, as long as there is a limit it converges, right?
infinity or -infinity means it diverges
if the limit is infinity then it will be divergence
infinity isn't a number just so you know it just means it gets really really big
but @freckles he said limit is 0 so i said he is correct
why is the limit 0 @Haseeb96
Ok. Well, thanks all. I need to go back and review I guess.
I'm pretty sure the product of really big positive numbers can not be 0
do you understand why it diverges @AmTran_Bus
Yes, I understand if it is infinity it diverges. Thanks. But I need to solve the limit correctly.
But I see what you are saying with that @freckles
\(a_n=n(n-6)\) \(\displaystyle \lim_{n\rightarrow\infty }n(n-6)\) \(\displaystyle \lim_{n\rightarrow\infty }n^2-6n\) diverges to positive inifinity.
Thanks solomon
Sure.... everytime to see the sequence convergence, for any sequence \(A_n\) , take the limit of it as n approaches infinity
60(54)=? 100(94)=? 1000(994)=? 10000(9994)=? these products are getting super super big
And if sequence diverges, then (always!) the series diverges
Yes. I super see that now. Thanks so much freckles. I think I understand it.
Thanks solomonzelman
yw
((that was my favorite section when I learned it, btw))
good luck, you will get an A I am sure... !

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