The Last Question Before My Graduation From College

Calculus Level 3

lim ⁡ n → ∞ 1 3 + 2 3 + ⋯ + n 3 n 3 − n 4 = ? \large\lim_{n\to\infty}\frac {1^3+2^3+\cdots+n^3}{n^3}-\frac n4=\, ?


The answer is 0.5.

Carwaniwer Qee by Carwaniwer Qee , 25, China

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2 solutions

Brian Lie
Jun 14, 2018

Relevant wiki: Stolz–Cesàro theorem

L = lim ⁡ n → ∞ 1 3 + 2 3 + ⋯ + n 3 n 3 − n 4 = lim ⁡ n → ∞ 4 ( 1 3 + 2 3 + ⋯ + n 3 ) − n 4 4 n 3 = lim ⁡ n → ∞ 4 ( n + 1 ) 3 − ( n + 1 ) 4 + n 4 4 ( n + 1 ) 3 − 4 n 3 = lim ⁡ n → ∞ 6 n 2 + 8 n + 3 12 n 2 + 12 n + 4 = 1 2 = 0.5 \begin{aligned} L&=\lim_{n\to\infty}\frac {1^3+2^3+\cdots+n^3}{n^3}-\frac n4 \\&=\lim_{n\to\infty}\frac {4(1^3+2^3+\cdots+n^3)-n^4}{4n^3} \\&=\lim_{n\to\infty}\frac {4(n+1)^3-(n+1)^4+n^4}{4(n+1)^3-4n^3} \\&=\lim_{n\to\infty}\frac {6n^2+8n+3}{12n^2+12n+4} \\&=\frac 12=\boxed{0.5} \end{aligned}

5 Helpful 1 Interesting 2 Brilliant 0 Confused
Naren Bhandari
Jun 14, 2018

L = lim ⁡ n → ∞ 1 3 + 2 3 + ⋯ + n 3 n 3 − n 4 = lim ⁡ n → ∞ n 2 ( n + 1 ) 2 4 n 3 − n 4 = lim ⁡ n → ∞ n 2 ( n + 1 ) 2 − n 4 4 n 3 = lim ⁡ n → ∞ n 4 + 2 n 3 + n 2 − n 4 4 n 3 = 1 2 L = \lim_{n\to \infty}\dfrac{1^3+2^3+\cdots +n^3}{n^3} -\dfrac{n}{4} =\lim_{n\to \infty} \dfrac{n^2(n+1)^2}{4n^3} -\dfrac{n}{4}= \lim_{n\to \infty} \dfrac{n^2(n+1)^2 - n^4}{4n^3} = \lim_{n\to \infty} \dfrac{n^4 +2n^3+n^2 - n^4}{4n^3} =\dfrac{1}{2}

6 Helpful 1 Interesting 0 Brilliant 0 Confused

You should carry the limit to infinity sign along until the last expression.

Marta Reece - 2 years, 12 months ago

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Oh!! I thought that I had taken unfortunately I hadn't. I have amended it .

Thank you !!

Naren Bhandari - 2 years, 12 months ago

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