What is the integer part of $\sum_{i=2}^{9999} \frac {1}{\sqrt i}?$

Question

What is the integer part of: $$\sum_{i=2}^{9999} \frac {1}{\sqrt {i}}$$

A short but tricky problem. Any help is welcome.

Source

Answer 1

Creative telescoping gives another chance. Since:

$$ 2\left(\sqrt{i+1}-\sqrt{i}\right)=\frac{2}{\sqrt{i+1}+\sqrt{i}}\leq\frac{1}{\sqrt{i}}\leq\frac{2}{\sqrt{i}+\sqrt{i-1}}=2\left(\sqrt{i}-\sqrt{i-1}\right) $$ we have that $\sum_{i=2}^{9999}\frac{1}{\sqrt{i}}$ is between $2\left(\sqrt{10000}-\sqrt{2}\right)$ and $2\left(\sqrt{9999}-1\right)$, hence: $$ \left\lfloor\sum_{i=2}^{9999}\frac{1}{\sqrt{i}}\right\rfloor = \color{red}{197}.$$

Answer 2

Using an approach similar to this, we have

$$1+\int_2^{10000}\frac 1{\sqrt i}di\quad<\quad\sum_{i=1}^{9999}\frac 1{\sqrt i} \quad < \quad 1+\int_1^{9999}\frac 1{\sqrt{i}}di\\ \int_2^{10000}\frac 1{\sqrt i}di\quad<\quad\sum_{i=2}^{9999}\frac 1{\sqrt i} \quad < \quad \int_1^{9999}\frac 1{\sqrt{i}}di\\ 2(\sqrt{10000}-\sqrt{2})\quad <\quad \sum_{i=2}^{9999}\frac 1{\sqrt i} \quad <\quad 2(\sqrt{9999}-\sqrt{1})\\ 197.17\quad <\quad \sum_{i=2}^{9999}\frac 1{\sqrt i} \quad <\quad 197.99\\ \qquad \qquad\qquad \quad \Biggr\lfloor{\sum_{i=2}^{9999}\frac 1{\sqrt i}}\Biggr\rfloor\quad = \quad 197\qquad\blacksquare $$