$$\frac{3}{4} \lim_{n \to \infty}\left(\frac{\sum_{r=1}^{n}\frac{1}{\sqrt{r}} \sum_{r=1}^{n}\sqrt{r} }{\sum_{r=1}^{n}r }\right)$$
Apparently, the answer is 2.
My try:
$$\frac{3}{4} \lim_{n \to \infty}\left(\frac{n^2}{\frac{n(n+1)}{2} }\right)$$
$$\frac{3}{2} \lim_{n\to \infty}\left(1-\frac{1}{n+1}\right)$$
Which is $\frac{3}{2}$. Surely I did something stupid or illegal here. What was it?
Also on the forum where I found this, It was said it can be done by converting into integrals, A push in the right direction on that too would be pretty rad.
Hope I am not asking too much.
Best Answer
Here is a solution using the suggested integral approach. Notice that \begin{align*} \frac{{\sum\nolimits_{r = 1}^n {\frac{1}{{\sqrt r }}} \sum\nolimits_{r = 1}^n {\sqrt r } }}{{\sum\nolimits_{r = 1}^n r }} = \frac{{\sum\nolimits_{r = 1}^n {\sqrt {\frac{n}{r}} } \sum\nolimits_{r = 1}^n {\sqrt {\frac{r}{n}} } }}{{\sum\nolimits_{r = 1}^n r }} & = \frac{{\frac{1}{{n^2 }}\sum\nolimits_{r = 1}^n {\sqrt {\frac{n}{r}} } \sum\nolimits_{r = 1}^n {\sqrt {\frac{r}{n}} } }}{{\frac{1}{{n^2 }}\sum\nolimits_{r = 1}^n r }} \\ &= \frac{{\left( {\frac{1}{n}\sum\nolimits_{r = 1}^n {\sqrt {\frac{n}{r}} } } \right)\left( {\frac{1}{n}\sum\nolimits_{r = 1}^n {\sqrt {\frac{r}{n}} } } \right)}}{{\frac{1}{n}\sum\nolimits_{r = 1}^n {\frac{r}{n}} }}. \end{align*} Thus, noticing the Riemann sums, we find \begin{align*} \mathop {\lim }\limits_{n \to + \infty } \frac{3}{4}\frac{{\sum\nolimits_{r = 1}^n {\frac{1}{{\sqrt r }}} \sum\nolimits_{r = 1}^n {\sqrt r } }}{{\sum\nolimits_{r = 1}^n r }} & = \frac{3}{4}\frac{{\mathop {\lim }\limits_{n \to + \infty } \left( {\frac{1}{n}\sum\nolimits_{r = 1}^n {\sqrt {\frac{n}{r}} } } \right)\mathop {\lim }\limits_{n \to + \infty } \left( {\frac{1}{n}\sum\nolimits_{r = 1}^n {\sqrt {\frac{r}{n}} } } \right)}}{{\mathop {\lim }\limits_{n \to + \infty } \frac{1}{n}\sum\nolimits_{r = 1}^n {\frac{r}{n}} }} \\ & = \frac{3}{4}\frac{{\int_0^1 {\frac{{dx}}{{\sqrt x }}} \int_0^1 {\sqrt x dx} }}{{\int_0^1 {xdx} }} = 2. \end{align*}