Can We Partition These Limits?

Let's multiply the expression by $\dfrac{\frac1{\sqrt n} \cdot \frac1{\sqrt n} }{\frac1n}$ , we have

$\dfrac{\left(\sqrt{ \frac1n} +\sqrt{ \frac2n} + \cdots +\sqrt { \frac nn} \right) \left( \frac1{\sqrt {1n}} + \frac1{\sqrt {2n}} + \cdots + \frac1{\sqrt {n\cdot n}} \right)}{\frac1n + \frac2n + \cdots + \frac nn}$

Let $A_n$ and $B_n$ denote the expressions, respectively, inside the first and second brackets in the numerator of the fraction above. Similarly, let $C_n$ denote the expression in the denominator of the fraction above.

So we want to evaluate $\displaystyle \lim_{n\to\infty} \dfrac{A_n B_n}{C_n}$ , and if the limits $A_n ,B_n, C_n$ are all finite values, then $\displaystyle \lim_{n\to\infty} \dfrac{A_n B_n}{C_n} = \displaystyle (\lim_{n\to\infty} A_n) \times (\lim_{n\to\infty}B_n) \div(\lim_{n\to\infty} C_n)$ is true.

Let's evaluate the limits of $A_n,B_n$ and $C_n$ separately.

Since $A_n, B_n$ and $C_n$ can be expressed in the form of $\displaystyle \dfrac1n \sum_{r=1}^\infty f\left( \dfrac rn\right)$ for some function $f(x)$ , then we can evaluate all these limits by Riemann sums .

Thus,
$\displaystyle \lim_{n\to\infty}A_n = \lim_{n\to\infty} \left(\sqrt{ \frac1n} +\sqrt{ \frac2n} + \cdots +\sqrt { \frac nn} \right) = \int_0^1 x^{1/2} \, dx = \dfrac1{1 + \frac12} x^{3/2} \bigg |_{x=0}^{x=1} = \dfrac23$ .
$\displaystyle \lim_{n\to\infty}B_n = \lim_{n\to\infty} \left( \frac1{\sqrt {1n}} + \frac1{\sqrt {2n}} + \cdots + \frac1{\sqrt {n\cdot n}} \right) = \int_0^1 x^{-1/2} \, dx = \dfrac1{1 - \frac12} x^{1/2} \bigg |_{x=0}^{x=1} = 2$ .
$\displaystyle \lim_{n\to\infty}C_n = \lim_{n\to\infty} \left( \frac1n + \frac2n + \cdots + \frac nn \right) = \int_0^1 x\, dx = \dfrac1{1 +1} x^{2} \bigg |_{x=0}^{x=1} = \dfrac12$ .

Since we have shown that the limits $A_n ,B_n, C_n$ are all non-zero finite values, then $\displaystyle \lim_{n\to\infty} \dfrac{A_n B_n}{C_n} = \displaystyle (\lim_{n\to\infty} A_n) \times (\lim_{n\to\infty}B_n) \div(\lim_{n\to\infty} C_n)$ is indeed true.

Hence our answer is just $\dfrac{\frac23 \cdot 2}{\frac12} =\boxed{{\dfrac83}}$ .