Limit of a sum of infinite series How do I find the following?
$$\lim_{n\to\infty} \frac{\left(\sum_{r=1}^n\sqrt{r}\right)\left(\sum_{r=1}^n\frac1{\sqrt{r}}\right)}{\sum_{r=1}^n r}$$ 
The lower sum is easy to find. However, I don't think there is an expression for the sums of the individual numerator terms... Nor can I think of a way to get the combined sum.
I just need a hint.

A way is to use Abel's summation to get $$\sum_{r\leq n}\sqrt{r}=\sum_{r\leq n}1\cdot\sqrt{r}=n^{3/2}-\frac{1}{2}\int_{1}^{n}\frac{\left\lfloor t\right\rfloor }{\sqrt{t}}dt
 $$ where $\left\lfloor t\right\rfloor 
 $ is the integer part of $t
 $ and using $\left\lfloor t\right\rfloor =t+O\left(1\right)
 $ we have $$\sum_{r\leq n}\sqrt{r}=\frac{2}{3}n^{3/2}+O\left(\sqrt{n}\right)
 $$ and, in a similar way $$\sum_{r\leq n}\frac{1}{\sqrt{r}}=\sqrt{n}+\frac{1}{2}\int_{1}^{n}\frac{\left\lfloor t\right\rfloor }{t^{3/2}}dt=\frac{3}{2}\sqrt{n}+O\left(1\right)
 $$ and for the last sum the well-known identity $$\sum_{r\leq n}r=\frac{n\left(n+1\right)}{2}.$$
