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I want to evaluate the following quotient limit:

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

I know an exact same answer is already in here.

I want to avoid using the fact that the integral is a limit of Riemann Sum, instead please refer to the following exercises that precede the evaluation of this limit.

  1. (a) Let $f$ be an increasing continuous function, defined for all $x \ge 1$, such that $f(x) \ge 0$. Show that

$$f(1) + f(2) + ... + f(n-1) \le \int_{1}^n f(x) dx \le f(2) + ... + f(n)$$

  1. (b) Let $F(x) = \int_{1}^x f(t) dt.$ Assume that $F(n) \to \infty$ as $n \to \infty$, and that

$$\lim_{n \to \infty} \frac{f(n)}{F(n)} = 0$$.

$\qquad$ Show that

$$\lim_{n \to \infty} \frac{f(1) + f(2) + ... + f(n)}{F(n)} = 1.$$

I have actually proved 1(a) and 1(b) as stated above. As with most exercises from Lang, usually the exercises are interconnected, so I would like to use 1(b) to show that the quotient limit tends to something.

I let $f(x) = x^{\frac{1}{3}}$. Of course this function satisfies all the assumptions in 1(a) and 1(b). Similarly we let $F(x) = \int_{1}^x t^{\frac{1}{3}}dt$. Now let's consider the limit again:

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

We see that the numerator of the quotient is $f(1) + f(2) + ... + f(n)$. Now consider $F(n) = \int_{1}^n t^{\frac{1}{3}}dt$ = $\frac{3}{4} (n^{\frac{4}{3}} - 1).$ But this is not the same as the denominator of the quotient, $n^{\frac{4}{3}}$. In this case how do I modify my solution so that I can actually use 1(b) as above?

Community
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Daniel
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    what you use is essentially asymptotic equivalence of the sum and corresponding integral, which is used in integral test for series convergence. If you don't like it for some reason use Euler-Maclaurin formula instead. – Alex Sep 28 '16 at 10:29

3 Answers3

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$$\lim_{n \to \infty} \frac{1^{1/3} + 2^{1/3} + \cdots + n^{1/3}}{n^{4/3}} = \lim_{n \to \infty} \frac{1^{1/3} + 2^{1/3} + \cdots + n^{1/3}}{\frac{3}{4} (n^{\frac{4}{3}} - 1)}\cdot\frac{\frac{3}{4} (n^{\frac{4}{3}} - 1)}{n^{4/3}}$$

First term goes to $1$ by your $\frac{f(1) + f(2) + ... + f(n)}{F(n)}\to1$ result, second term goes to $3/4$ by standard limits.

So, your final answer is $3/4$.

πr8
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  • SIgh I cannot imagine I actually did not think of this easy trick...cry. Thanks, will accept your answer later! – Daniel Sep 28 '16 at 10:24
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Using generalized harmonic numbers $$\sum_{i=1}^n i^{\frac 1 3}=H_n^{\left(-\frac{1}{3}\right)}$$ which make $$S_n=\frac{\sum_{i=1}^n i^{\frac 1 3}}{n^{\frac 4 3}}=\frac{H_n^{\left(-\frac{1}{3}\right)}}{n^{4/3}}$$ Now, using the asymptotics $$S_n=\frac{3}{4}+\frac{1}{2 n}+O\left(\frac{1}{n^{4/3}}\right)$$ which shows the limit and how it is approached.

Claude Leibovici
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  • Thanks for the input, I am not familiar with the generalized harmonic numbers and asymptotics yet, but I will definitely take note going forward :) – Daniel Sep 28 '16 at 12:49
  • @Daniel. So, don't worry : consider that this was for your curiosity and remember when they will teach you all that stuff. Cheers and good luck in your studies. :-) – Claude Leibovici Sep 28 '16 at 16:34
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You may like this solution. By the Stolz-Cesaro theorem, \begin{eqnarray} &&\lim_{n \to \infty} \frac{1^{1/3} + 2^{1/3} + \cdots + n^{1/3}}{n^{4/3}}\\ &=&\lim_{n \to \infty} \frac{(n+1)^{1/3}}{(n+1)^{4/3}-n^{4/3}}\\ &=&\lim_{n \to \infty} \frac{(n+1)^{1/3}}{[(n+1)^{1/3}-n^{1/3}][(n+1)^{1/3}+n^{1/3}][(n+1)^{2/3}+n^{2/3}]}\\ &=&\lim_{n \to \infty} \frac{(n+1)^{1/3}[(n+1)^{2/3}+n^{1/3}(n+1)^{1/3}+n^{2/3}]}{[(n+1)^{1/3}+n^{1/3}][(n+1)^{2/3}+n^{2/3}]}\\ &=&\lim_{n \to \infty} \frac{1+\left(\frac{n}{n+1}\right)^{1/3}+\left(\frac{n}{n+1}\right)^{2/3}}{[1+\left(\frac{n}{n+1}\right)^{1/3}][1+\left(\frac{n}{n+1}\right)^{2/3}]}\\ &=&\frac34. \end{eqnarray}

xpaul
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