anonymous
  • anonymous
how do i find the limit as an goes to infinity of the following? (the equation's in the comments)
Mathematics
katieb
  • katieb
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

anonymous
  • anonymous
\[\frac{ 4 }{ n }\sum_{a=1}^{n}(-(\frac{ 4a }{ n })^{2} -4(\frac{ 4a }{ n }))\] This is the equation in question
anonymous
  • anonymous
\[\frac{ 4 }{ n }\sum_{a=1}^{n}(-(\frac{ 4a }{ n })^{2}) - \frac{ 4 }{ n }\sum_{a=1}^{n}(4(\frac{ 4a }{ n}))\]
anonymous
  • anonymous
we're not sure where to go from here

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
looks like you are trying to compute some sort or reimann sum
anonymous
  • anonymous
exactly. a right riemann sum of -x^2 -4x to be exact.
anonymous
  • anonymous
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 \[\sum_{a=1}^na\]
tkhunny
  • tkhunny
After a little algebra. \(-\dfrac{16}{n^{2}}\left[\dfrac{1}{n}\sum\limits_{a=1}^{n}a^{2} + \sum\limits_{a=1}^{n}a\right]\)
primeralph
  • primeralph
@satellite73 Got this?
anonymous
  • anonymous
\[\frac{ 4 }{ n }\sum_{a=1}^{n}(-(\frac{ 4a }{ n })^{2})\] \[=-\frac{64}{n^3}\sum_{a=1}^na^2\] for example @primeralph maybe, i am tired and all this latex is wearing me out
anonymous
  • anonymous
@tkhunny i think the \(4\) is being squared as well
tkhunny
  • tkhunny
Whoops. Make that 64, instead of 16.
anonymous
  • anonymous
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
anonymous
  • anonymous
Wait but how does this help us get the limit as n -> infinity?
anonymous
  • anonymous
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
anonymous
  • anonymous
as you remember from some pre calc course about "horizontal asymptotes" that makes the limit \(-\frac{128}{6}\)
anonymous
  • anonymous
oh oops duh that makes sense.
anonymous
  • anonymous
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
anonymous
  • anonymous
Thanks so much!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
anonymous
  • anonymous
@satellite73 where did -128 come from?
anonymous
  • anonymous
sorry probably a dumb question but...
anonymous
  • anonymous
oh waitn.....nevermind

Looking for something else?

Not the answer you are looking for? Search for more explanations.