What is the integer part of $\sum_{i=2}^{9999} \frac {1}{\sqrt i}?$ What is the integer part of: $$\sum_{i=2}^{9999} \frac {1}{\sqrt {i}}$$
A short but tricky problem. Any help is welcome.
Source
 A: 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:
$$ \left\lfloor\sum_{i=2}^{9999}\frac{1}{\sqrt{i}}\right\rfloor = \color{red}{197}.$$
A: Using an approach similar to this, we have
$$1+\int_2^{10000}\frac 1{\sqrt i}di\quad<\quad\sum_{i=1}^{9999}\frac 1{\sqrt i} \quad < \quad 1+\int_1^{9999}\frac 1{\sqrt{i}}di\\
\int_2^{10000}\frac 1{\sqrt i}di\quad<\quad\sum_{i=2}^{9999}\frac 1{\sqrt i} \quad < \quad \int_1^{9999}\frac 1{\sqrt{i}}di\\
2(\sqrt{10000}-\sqrt{2})\quad <\quad \sum_{i=2}^{9999}\frac 1{\sqrt i} \quad <\quad 2(\sqrt{9999}-\sqrt{1})\\
197.17\quad <\quad \sum_{i=2}^{9999}\frac 1{\sqrt i} \quad <\quad 197.99\\
\qquad \qquad\qquad \quad \Biggr\lfloor{\sum_{i=2}^{9999}\frac 1{\sqrt i}}\Biggr\rfloor\quad = \quad 197\qquad\blacksquare
$$
