Sum of Series 1/√r

Question

What will a good approximation of the following series ?

S=Σ(1/√r) where r varies from 1 to N=10⁶ Specifically, we need to find floor(S)

I tried this by two methods,

1.Using the inequality, S<2√N , which can be proven easily by induction. Using this I got S<2000 but wasn't able to find out floor(S) as S can be 1999 or 1998 or anything below, but close to 2000.

2. Replacing terms adjacent to perfect squares by perfect squares and adjusting the inequality , i.e replacing √3 by √4 and so on. This was a huge failure as it couldn't work for large values of N.

How would we obtain the right value? Is there a general method for any value of N?

Answer 1

We could estimate this sum by using an integral over $f(x)=\frac{1}{\sqrt{x}}$ since $f(x)$ is positive and decreasing over $[1,\infty)$. So

$$\sum_{n=1}^k \frac{1}{\sqrt{n}}\approx\int_1^k \frac{dx}{\sqrt{x}}=2\sqrt{x}|_{x=1}^k=2(\sqrt{k}-1)$$ For $k=10^6$, $$\sum_{n=1}^{10^6}\frac{1}{\sqrt{n}}\approx2(10^3-1)=1998$$

Answer 2

We can easily prove

$$2(\sqrt{r + 1} - \sqrt{r}) < \frac{1}{\sqrt{r}} < 2(\sqrt{r} - \sqrt{r- 1})$$

So $$2\sum^{10^{6}}_{r=1}\bigg[\sqrt{r+1}-\sqrt{r}\bigg]<1+\sum^{10^{6}}_{r=2}\frac{1}{\sqrt{r}}<1+2\sum^{10^6}_{r=2}\bigg[\sqrt{r}-\sqrt{r-1}\bigg]$$

So $$2\bigg(1000.01-1\bigg)<\sum^{10^6}_{r=1}\frac{1}{\sqrt{r}}<2(10^3-1)+1$$

So we have $$\Bigg\lfloor \sum^{10^6}_{r=1}\frac{1}{\sqrt{r}}\Bigg\rfloor =1998$$

$\text{Added:}$ For $r>1$

$$\begin{gathered} \frac{1}{{\sqrt r }} = \frac{2}{{\sqrt r + \sqrt r }} < \frac{2}{{\sqrt r + \sqrt {r - 1} }} = 2(\sqrt r - \sqrt {r - 1} ) \hfill \\ \hfill \\ \hfill \\ \frac{1}{{\sqrt r }} = \frac{2}{{\sqrt r + \sqrt r }} > \frac{2}{{\sqrt {r + 1} + \sqrt r }} = 2(\sqrt {r + 1} - \sqrt r ) \hfill \\ \end{gathered}$$