Limit of a sum of infinite series

Question

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.

Answer 1

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

Answer 2

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:

$$\lim_{n\to\infty}\frac{2\text{H}_n^{-\frac{1}{2}}\cdot\text{H}_n^{\frac{1}{2}}}{n(1+n)}=2\lim_{n\to\infty}\frac{\text{H}_n^{-\frac{1}{2}}\cdot\text{H}_n^{\frac{1}{2}}}{n(1+n)}=$$ $$2\lim_{n\to\infty}\frac{\frac{\text{d}}{\text{d}n}\left(\text{H}_n^{-\frac{1}{2}}\cdot\text{H}_n^{\frac{1}{2}}\right)}{\frac{\text{d}}{\text{d}n}n(1+n)}=2\lim_{n\to\infty}\frac{\frac{1}{2}\left(\text{H}_n^{-\frac{1}{2}}\zeta\left(\frac{3}{2},n+1\right)-\text{H}_n^{\frac{1}{2}}\zeta\left(\frac{1}{2},n+1\right)\right)}{2n+1}=$$ $$\lim_{n\to\infty}\frac{\text{H}_n^{-\frac{1}{2}}\zeta\left(\frac{3}{2},n+1\right)-\text{H}_n^{\frac{1}{2}}\zeta\left(\frac{1}{2},n+1\right)}{2n+1}=\lim_{n\to\infty}\frac{\frac{\text{d}}{\text{d}n}\left(\text{H}_n^{-\frac{1}{2}}\zeta\left(\frac{3}{2},n+1\right)-\text{H}_n^{\frac{1}{2}}\zeta\left(\frac{1}{2},n+1\right)\right)}{\frac{\text{d}}{\text{d}n}\left(2n+1\right)}=$$ $$\lim_{n\to\infty}\frac{\frac{1}{2}\left(3\zeta\left(\frac{5}{2},1+n\right)\left(\zeta\left(-\frac{1}{2},1+n\right)-\zeta\left(-\frac{1}{2}\right)\right)+\zeta\left(\frac{3}{2},1+n\right)\left(-3\zeta\left(\frac{1}{2},1+n\right)+\zeta\left(\frac{1}{2}\right)\right)\right)}{2}=$$ $$\frac{1}{4}\lim_{n\to\infty}\left(3\zeta\left(\frac{5}{2},1+n\right)\left(\zeta\left(-\frac{1}{2},1+n\right)-\zeta\left(-\frac{1}{2}\right)\right)+\zeta\left(\frac{3}{2},1+n\right)\left(-3\zeta\left(\frac{1}{2},1+n\right)+\zeta\left(\frac{1}{2}\right)\right)\right)$$

Answer 3

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 $$\sum_{r\leq n}r=\frac{n\left(n+1\right)}{2}.$$