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I tried to evaluate this limit:

$$\lim_{n\to \infty} \frac {1 + \sqrt[3]{2} + \sqrt[3]{3} +\cdots+ \sqrt[3]{n}}{n^{4/3}}$$

My work:

I was just following suit in the book I read, so I came up with this solution to get the limit above:

$$\int_0 ^1 \sqrt[3]{x} \, dx = \frac{3}{4}$$

The problem is, the way I solved the problem above, I did not think what is the signficance of the denominator, which is $n^{\frac{4}{3}}$. I might have the same solution to the problem, as long as i see the $\sqrt[3]{n}$ at the end, not minding the denominator.

What is the correct approach in getting the limit above?

Michael Hardy
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fitzmerl duron
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1 Answers1

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The sum can be written $$\frac 1n\sum_{k=1}^n\sqrt[3]{\frac kn}$$ which is a Riemann sum for the function $f(x)=\sqrt[3]x$ on the interval $[0,1]$.

ajotatxe
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  • I must remember that form......the coefficient is $\frac{1}{n}$ and $\frac{k}{n}$ becomes $x$ always.... – fitzmerl duron Aug 24 '17 at 16:17
  • Wait...the original form is $\frac {1}{n^\frac{4}{3}}\sum_{k=1}^n\sqrt[3]{\frac kn}$ ? and when modified, it becomes $\frac 1n\sum_{k=1}^n\sqrt[3]{\frac kn}$? how did you modified it? from $\frac {1}{n^\frac{4}{3}}\sum_{k=1}^n\sqrt[3]{\frac kn}$ to $\frac 1n\sum_{k=1}^n\sqrt[3]{\frac kn}$? – fitzmerl duron Aug 26 '17 at 16:06