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: As a rough approximation,
$$\sum_{r=1}^n\sqrt{r}\approx\int_1^{n+1}\sqrt{r}dr$$
A: HINTS:


*

*Summations:
$$\sum_{r=1}^{n}\sqrt{r}=\text{H}_n^{-\frac{1}{2}}$$
$$\sum_{r=1}^{n}\frac{1}{\sqrt{r}}=\text{H}_n^{\frac{1}{2}}$$
$$\sum_{r=1}^{n}r=\frac{1}{2}n(1+n)$$


With $\text{H}_n^r$ is the generalized harmonic number.


*

*Fraction:
$$\frac{\text{H}_n^{-\frac{1}{2}}\cdot\text{H}_n^{\frac{1}{2}}}{\frac{1}{2}n(1+n)}=\frac{2\text{H}_n^{-\frac{1}{2}}\cdot\text{H}_n^{\frac{1}{2}}}{n(1+n)}$$

*Limit:
$$\lim_{n\to\infty}\frac{2\text{H}_n^{-\frac{1}{2}}\cdot\text{H}_n^{\frac{1}{2}}}{n(1+n)}=2\lim_{n\to\infty}\frac{\text{H}_n^{-\frac{1}{2}}\cdot\text{H}_n^{\frac{1}{2}}}{n(1+n)}=$$
$$2\lim_{n\to\infty}\frac{\frac{\text{d}}{\text{d}n}\left(\text{H}_n^{-\frac{1}{2}}\cdot\text{H}_n^{\frac{1}{2}}\right)}{\frac{\text{d}}{\text{d}n}n(1+n)}=2\lim_{n\to\infty}\frac{\frac{1}{2}\left(\text{H}_n^{-\frac{1}{2}}\zeta\left(\frac{3}{2},n+1\right)-\text{H}_n^{\frac{1}{2}}\zeta\left(\frac{1}{2},n+1\right)\right)}{2n+1}=$$
$$\lim_{n\to\infty}\frac{\text{H}_n^{-\frac{1}{2}}\zeta\left(\frac{3}{2},n+1\right)-\text{H}_n^{\frac{1}{2}}\zeta\left(\frac{1}{2},n+1\right)}{2n+1}=\lim_{n\to\infty}\frac{\frac{\text{d}}{\text{d}n}\left(\text{H}_n^{-\frac{1}{2}}\zeta\left(\frac{3}{2},n+1\right)-\text{H}_n^{\frac{1}{2}}\zeta\left(\frac{1}{2},n+1\right)\right)}{\frac{\text{d}}{\text{d}n}\left(2n+1\right)}=$$
$$\lim_{n\to\infty}\frac{\frac{1}{2}\left(3\zeta\left(\frac{5}{2},1+n\right)\left(\zeta\left(-\frac{1}{2},1+n\right)-\zeta\left(-\frac{1}{2}\right)\right)+\zeta\left(\frac{3}{2},1+n\right)\left(-3\zeta\left(\frac{1}{2},1+n\right)+\zeta\left(\frac{1}{2}\right)\right)\right)}{2}=$$
$$\frac{1}{4}\lim_{n\to\infty}\left(3\zeta\left(\frac{5}{2},1+n\right)\left(\zeta\left(-\frac{1}{2},1+n\right)-\zeta\left(-\frac{1}{2}\right)\right)+\zeta\left(\frac{3}{2},1+n\right)\left(-3\zeta\left(\frac{1}{2},1+n\right)+\zeta\left(\frac{1}{2}\right)\right)\right)$$
A: 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 $$\sum_{r\leq n}r=\frac{n\left(n+1\right)}{2}.$$
