# calculus

Evaluate

lim (1³ +2³ +3³ +…+ n3)/n^4
n →∞

by showing that the limit is a particular definite integral and evaluating that definite integral.

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1. sum(1) = 1^3/1^4 = 1 or 4/4
sum(2) = (1 + 8)/16 = 9/16
sum(3) = (1+8+27)/81 = 36/81
= 4/9 or 16/36
sum(4) = (1+8+27+64)/256
= (1+8+27+64)/256
= 100/256 = 25/64
sum(5) = (1+8+27+64+125)/625
= 225/625 = 9/25 = 36/100

I noticed I can change my numerator to (n+1)^2 for sum(n)
and the denominators are
1,16,36,64,100
or 4(1^2), 4(2^2), 4(3^2), 4(5^2)

ahhh, so we can say:
(1³ +2³ +3³ +…+ n3)/n^4 = (n+1)^2/(4n^2)

you might want to check this for one or more sums

so lim(1³ +2³ +3³ +…+ n3)/n^4 as x --->∞
= lim (n+1)^2 / (4n^2) as x ---> ∞
= lim (n^2 + 2n + 1)/4n^2 as x ---> ∞
= lim ( 1/4 + 1/(2n) + 1/(4n^2) as x ---> ∞

= 1/4

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posted by Reiny

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