$\large S = 1 + \frac 1{\sqrt 2} + \frac 1{\sqrt 3} + \frac 1{\sqrt 4} + \cdots + \frac 1{\sqrt{99}} + \frac 1{\sqrt{100}}$

Find $\lfloor S \rfloor$ .

**
Notation:
**
$\lfloor \cdot \rfloor$
denotes
floor function
.

Other problems: Check your Calibre

The answer is 18.

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Use induction to prove that $\sqrt {k+1}$ - $\sqrt {k}$ < $\frac{1}{2* \sqrt {k}}$ < $\sqrt {k}$ - $\sqrt {k-1}$ ,Then proceed you will easily get the answer by telescoping series.

:).