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

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alffer1 Group TitleBest ResponseYou've already chosen the best response.0
\[\frac{ 4 }{ n }\sum_{a=1}^{n}((\frac{ 4a }{ n })^{2} 4(\frac{ 4a }{ n }))\] This is the equation in question
 11 months ago

alffer1 Group TitleBest ResponseYou've already chosen the best response.0
\[\frac{ 4 }{ n }\sum_{a=1}^{n}((\frac{ 4a }{ n })^{2})  \frac{ 4 }{ n }\sum_{a=1}^{n}(4(\frac{ 4a }{ n}))\]
 11 months ago

alffer1 Group TitleBest ResponseYou've already chosen the best response.0
we're not sure where to go from here
 11 months ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.2
looks like you are trying to compute some sort or reimann sum
 11 months ago

alffer1 Group TitleBest ResponseYou've already chosen the best response.0
exactly. a right riemann sum of x^2 4x to be exact.
 11 months ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.2
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 \[\sum_{a=1}^na\]
 11 months ago

tkhunny Group TitleBest ResponseYou've already chosen the best response.0
After a little algebra. \(\dfrac{16}{n^{2}}\left[\dfrac{1}{n}\sum\limits_{a=1}^{n}a^{2} + \sum\limits_{a=1}^{n}a\right]\)
 11 months ago

primeralph Group TitleBest ResponseYou've already chosen the best response.0
@satellite73 Got this?
 11 months ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.2
\[\frac{ 4 }{ n }\sum_{a=1}^{n}((\frac{ 4a }{ n })^{2})\] \[=\frac{64}{n^3}\sum_{a=1}^na^2\] for example @primeralph maybe, i am tired and all this latex is wearing me out
 11 months ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.2
@tkhunny i think the \(4\) is being squared as well
 11 months ago

tkhunny Group TitleBest ResponseYou've already chosen the best response.0
Whoops. Make that 64, instead of 16.
 11 months ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.2
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
 11 months ago

alffer1 Group TitleBest ResponseYou've already chosen the best response.0
Wait but how does this help us get the limit as n > infinity?
 11 months ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.2
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
 11 months ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.2
as you remember from some pre calc course about "horizontal asymptotes" that makes the limit \(\frac{128}{6}\)
 11 months ago

alffer1 Group TitleBest ResponseYou've already chosen the best response.0
oh oops duh that makes sense.
 11 months ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.2
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
 11 months ago

aef321 Group TitleBest ResponseYou've already chosen the best response.0
Thanks so much!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
 11 months ago

alffer1 Group TitleBest ResponseYou've already chosen the best response.0
@satellite73 where did 128 come from?
 11 months ago

alffer1 Group TitleBest ResponseYou've already chosen the best response.0
sorry probably a dumb question but...
 11 months ago

alffer1 Group TitleBest ResponseYou've already chosen the best response.0
oh waitn.....nevermind
 11 months ago
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