TITLE: What is the integer part of $\sum_{i=2}^{9999} \frac {1}{\sqrt i}?$
QUESTION [6 upvotes]: What is the integer part of: $$\sum_{i=2}^{9999} \frac {1}{\sqrt {i}}$$
A short but tricky problem. Any help is welcome.
Source

REPLY [25 votes]: 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:
$$ \left\lfloor\sum_{i=2}^{9999}\frac{1}{\sqrt{i}}\right\rfloor = \color{red}{197}.$$