What is the value of the floor function challenge?

[anemone]

Find $$\left\lfloor{S}\right\rfloor$$ if $$S=1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\cdots+\frac{1}{\sqrt{80}}$$.

 

[kaliprasad]

Find $$\left\lfloor{S}\right\rfloor$$ if $$S=1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\cdots+\frac{1}{\sqrt{80}}$$.

Spoiler

we have $\frac{1}{\sqrt{k}} = \frac{2}{2\sqrt{k}} > \frac{2}{(\sqrt{k} + \sqrt{k +1} )} = 2(\sqrt{k+1} - \sqrt{k})$
hence adding above from k 1 to 80 we get $S > 2 * (\sqrt(81) -1) = 16$
also
$\frac{1}{\sqrt{k}} = \frac{2}{2\sqrt{k}} < \frac{2}{\sqrt{k} + \sqrt{k - 1 }} = 2(\sqrt{k } - \sqrt{k-1})$
adding above from 1 to 80 we get $S < 2\sqrt{80}$ and $\sqrt{80} > 8.5$ because $8.5^2= 72.25$ so $S < 17$
so integral part is 16

 

[anemone]

Spoiler

we have $\frac{1}{\sqrt{k}} = \frac{2}{2\sqrt{k}} > \frac{2}{(\sqrt{k} + \sqrt{k +1} )} = 2(\sqrt{k+1} - \sqrt{k})$
hence adding above from k 1 to 80 we get $S > 2 * (\sqrt(81) -1) = 16$
also
$\frac{1}{\sqrt{k}} = \frac{2}{2\sqrt{k}} < \frac{2}{\sqrt{k} + \sqrt{k - 1 }} = 2(\sqrt{k } - \sqrt{k-1})$
adding above from 1 to 80 we get $S < 2\sqrt{80}$ and $\sqrt{80} > 8.5$ because $8.5^2= 72.25$ so $S < 17$
so integral part is 16

Well done, kaliprasad!(Cool)