## aef321 Group Title how do i find the limit as an goes to infinity of the following? (the equation's in the comments) 8 months ago 8 months ago

1. alffer1 Group Title

$\frac{ 4 }{ n }\sum_{a=1}^{n}(-(\frac{ 4a }{ n })^{2} -4(\frac{ 4a }{ n }))$ This is the equation in question

2. alffer1 Group Title

$\frac{ 4 }{ n }\sum_{a=1}^{n}(-(\frac{ 4a }{ n })^{2}) - \frac{ 4 }{ n }\sum_{a=1}^{n}(4(\frac{ 4a }{ n}))$

3. alffer1 Group Title

we're not sure where to go from here

4. satellite73 Group Title

looks like you are trying to compute some sort or reimann sum

5. alffer1 Group Title

exactly. a right riemann sum of -x^2 -4x to be exact.

6. satellite73 Group Title

don't let the $$n$$ through you off in the summation you are summing over $$a$$ (never seen an $$a$$ used like that before, but no matter) pull it right out front of the sum then use the formula for $\sum_{a=1}^na^2$ and also $\sum_{a=1}^na$

7. tkhunny Group Title

After a little algebra. $$-\dfrac{16}{n^{2}}\left[\dfrac{1}{n}\sum\limits_{a=1}^{n}a^{2} + \sum\limits_{a=1}^{n}a\right]$$

8. primeralph Group Title

@satellite73 Got this?

9. satellite73 Group Title

$\frac{ 4 }{ n }\sum_{a=1}^{n}(-(\frac{ 4a }{ n })^{2})$ $=-\frac{64}{n^3}\sum_{a=1}^na^2$ for example @primeralph maybe, i am tired and all this latex is wearing me out

10. satellite73 Group Title

@tkhunny i think the $$4$$ is being squared as well

11. tkhunny Group Title

Whoops. Make that 64, instead of 16.

12. satellite73 Group Title

not really sure the sum is correct, but assuming it is for the moment $-\frac{64}{n^3}\sum_{k=1}^nk^2=-\frac{64}{n^3}\frac{n(n+1)(2n+1)}{6}$ for the first one

13. alffer1 Group Title

Wait but how does this help us get the limit as n -> infinity?

14. satellite73 Group Title

taking the limit as $$n\to \infty$$ is very easy because the numerator is a polynomial of degree 3 will leading coefficient -128 and the denominator is a polynomial of degree 3 with leading coefficent 6

15. satellite73 Group Title

as you remember from some pre calc course about "horizontal asymptotes" that makes the limit $$-\frac{128}{6}$$

16. alffer1 Group Title

oh oops duh that makes sense.

17. satellite73 Group Title

easy if you think about it that way instead of the way they do it in the text by dividing top and bottom by $$n^3$$ and all that other nonsense

18. aef321 Group Title

Thanks so much!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

19. alffer1 Group Title

@satellite73 where did -128 come from?

20. alffer1 Group Title

sorry probably a dumb question but...

21. alffer1 Group Title

oh waitn.....nevermind