Find the integer part of the sum $S=\sum_{k=1}^{80} \frac{1}{\sqrt k} $

Question

Let $$S=\sum_{k=1}^{80} \frac{1}{\sqrt k}.$$Then I would like to obtain $\lfloor S \rfloor$, the integer part of $S$.

I am not able to think how to start question .

Answer 1

An elementary way.

One may observe that, for $k=1,2,\cdots$, $$ \sqrt{k+1}-\sqrt{k}=\frac1{\sqrt{k+1}+\sqrt{k}},\quad \sqrt{k+1/2}-\sqrt{k-1/2}=\frac1{\sqrt{k+1/2}+\sqrt{k-1/2}} $$ and that $$ \frac1{\sqrt{k+1}+\sqrt{k}} <\frac1{2\sqrt{k}}< \frac1{\sqrt{k+\frac12}+\sqrt{k-\frac12}} $$ giving, for $k=1,2,\cdots$, $$ 2\left(\sqrt{k+1}-\sqrt{k}\right)<\frac1{\sqrt{k}}< 2\left(\sqrt{k+\frac12}-\sqrt{k-\frac12}\right) $$

then one may conclude with telescoping sums:

$$ 2\sqrt{n+1}-2<\sum_{k=1}^n\frac1{\sqrt{k}}< 2\sqrt{n+\frac12}-\sqrt{2},\quad n\ge1. $$

Taking $n=80$ gives $$ \color{blue}{16}<\sum_{k=1}^{80}\frac1{\sqrt{k}}<\color{blue}{16}.5301\cdots. $$

One may observe that $$ \left|\left(2\sqrt{n+\frac12}-\sqrt{2}\right)-(2\sqrt{n+1}-2)\right|<2-\sqrt{2}<1. $$

Answer 2

Using Abel's summation we get $$S\left(N\right)=\sum_{k=1}^{N}\frac{1}{\sqrt{k}}=\sqrt{N}+\frac{1}{2}\int_{1}^{N}\frac{\left\lfloor t\right\rfloor }{t^{3/2}}dt $$ so, using the bounds $t-1\leq\left\lfloor t\right\rfloor \leq t $ we have $$2\sqrt{N}+\frac{1}{\sqrt{N}}-2\leq S(N)\leq2\sqrt{N}-1 $$ hence $$\color{red}{\left\lfloor S\left(80\right)\right\rfloor =16.}$$