Evaluate the limit (1^3+2^3+...+n^3)/n^4. n-->infinite.

Asked on by maisaphie

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neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

Posted on

To find the lt (1^3+2^3+..n^3 )/n^4 as n--> infinite.

We know that 1^3+2^3+3^+...n^3 = (1/4){n(n+1)}^2.

Therefore Lt (1^3+2^3+3^3+4^3+....+(n-1)^3+n^3)/n^4 = lt (1/4){n(n+1)}^2 /n^4 as n--> infinity.  Both numerator and denominator become infinite.

So RHS = Lt (1/4) {n(n+1)}^2 / n^4 = Lt  (1/4){n^4+2n^3+n^2}/n^4 = Lt (1/4){1+2/n+1/n^2 as n-->infinity

= (1/4) {1+2*0+1*0} as 2/n and 1/n^2 approach zero as n--> ifinity.

= 1/4.

Therefore Lt (1^3+2^3+3^3+4^3+...+(n-1)^3+n^3}/n^2 = 1/4 as n-->infinity.

william1941's profile pic

william1941 | College Teacher | (Level 3) Valedictorian

Posted on

In this problem we see that the numerator

1^3+2^3+...n^3= [n*(n+1)/2]^2

= [n^2*(n+1)^2]/4

The denominator is n^4.

Therefore the expression simplified is: [n^2*(n+1)^2]/4*n^4




Now when n-->infinity (1/n)-->0

=>(1/4)[1+1/n^2+2/n] for n-->infinity



Therefore evaluating the required limit we get 1/4.

Top Answer

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

Before starting the evaluation of the limit, we'll re-write the sum of the numerator:

1^3+2^3+...+n^3 = [n*(n+1)/2]^2

To evaluate the limit of the rational function, when n tends to +inf.,we'll factorize both, numerator and denominator.

We'll substitute the numerator, by the result of the sum, we'll factorize  by the highest power of n, which in this case is n^4.

We'll have:

lim (1^3+2^3+...+n^3)/n^4 = lim n^2*(n+1)^2/4*lim n^4

lim (1^3+2^3+...+n^3)/n^4 = lim n^4(1/n^2+2/n+1)/4*lim n^4

We'll divide by n^4:

lim (1/n^2+2/n+1)/4*lim 1 = (0+0+1)/4*1

lim (1^3+2^3+...+n^3)/n^4 = 1/4

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