The limit of $\frac{1^{1/3} + 2^{1/3} + \cdots + n^{1/3}}{n^{4/3}} $ 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.


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


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


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