Here's the question you clicked on:
aef321
how do i find the limit as an goes to infinity of the following? (the equation's in the comments)
\[\frac{ 4 }{ n }\sum_{a=1}^{n}(-(\frac{ 4a }{ n })^{2} -4(\frac{ 4a }{ n }))\] This is the equation in question
\[\frac{ 4 }{ n }\sum_{a=1}^{n}(-(\frac{ 4a }{ n })^{2}) - \frac{ 4 }{ n }\sum_{a=1}^{n}(4(\frac{ 4a }{ n}))\]
we're not sure where to go from here
looks like you are trying to compute some sort or reimann sum
exactly. a right riemann sum of -x^2 -4x to be exact.
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 \[\sum_{a=1}^na\]
After a little algebra. \(-\dfrac{16}{n^{2}}\left[\dfrac{1}{n}\sum\limits_{a=1}^{n}a^{2} + \sum\limits_{a=1}^{n}a\right]\)
@satellite73 Got this?
\[\frac{ 4 }{ n }\sum_{a=1}^{n}(-(\frac{ 4a }{ n })^{2})\] \[=-\frac{64}{n^3}\sum_{a=1}^na^2\] for example @primeralph maybe, i am tired and all this latex is wearing me out
@tkhunny i think the \(4\) is being squared as well
Whoops. Make that 64, instead of 16.
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
Wait but how does this help us get the limit as n -> infinity?
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
as you remember from some pre calc course about "horizontal asymptotes" that makes the limit \(-\frac{128}{6}\)
oh oops duh that makes sense.
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
Thanks so much!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
@satellite73 where did -128 come from?
sorry probably a dumb question but...