Find the limit of S(n) as n- infinity
s(n) = ∑_(i=1)^n▒(8i-n)/n^2
I keep getting 3 which is not one of the available multiple choice answers. So if someone could double check this I would appreciate it.

Question

Find the limit of S(n) as n-> infinity 
s(n) = ∑_(i=1)^n▒(8i-n)/n^2 
I keep getting 3 which is not one of the available multiple choice answers.  So if someone could double check this I would appreciate it.

anonymous · Answer

hmm Microsoft word didnt copy correctly 
The equation is $$s(n) = \sum_{n}^{i=1}(8i-n)/n^2$$

anonymous · Answer

does it converge?

anonymous · Answer

I'm not entirely sure what you mean...this is a calc one intro to integration problem...

anonymous · Answer

I don't think it has a limit. I think it spins off to infinity (and probably beyond)

anonymous · Answer

and I didnt vet that properly I have my index flipped

anonymous · Answer

yes, but it doesn't matter too much

anonymous · Answer

oh ok :(

anonymous · Answer

what you are really asking (I think) is:
$$\sum_{n=1}^{\infty}\sum_{i=1}^{n}{8i-n \over n^2}$$

anonymous · Answer

your notation is not something I am familiar with but if it means that you are simply substituting $$\infty$$ for n then yes

anonymous · Answer

well that thing does not converge...

anonymous · Answer

$$(1/n^2)\sum_{i=1}^{n}8i-n$$