The value of the floor function for the series $$S=1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\cdots+\frac{1}{\sqrt{80}}$$ is determined to be $$\left\lfloor{S}\right\rfloor$$. The series converges to a numerical value that can be approximated using numerical methods or calculus techniques. The discussion highlights the importance of understanding series and convergence in mathematical analysis.
PREREQUISITESMathematicians, students studying calculus, and anyone interested in series convergence and numerical analysis.
anemone said:Find $$\left\lfloor{S}\right\rfloor$$ if $$S=1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\cdots+\frac{1}{\sqrt{80}}$$.
kaliprasad said: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