Find floor of sum $\sum_{k=1}^{80} k^{-1/2}$ We have to find the floor $\lfloor S \rfloor$ of the following sum:
$$S = \sum_{k=1}^{80}\frac{1}{\sqrt k}$$
What I  did was to find a approximate series that this series is near to. Let that series have general term $T_k$ and original series may have general term $a_k$. We construct the following series of $T_k$
$$T_k = \frac{1}{ \sqrt{k+1}+\sqrt{k}} = \sqrt{k+1}-\sqrt{k}\\$$
Then we have the following inequality:
$$\frac{a_k}{2} = \frac{1}{\sqrt{k}+\sqrt{k}} > T_k \\
\sum a_k > 2 \sum_{1}^{80} T_k \\
S > 2 (\sqrt{81}-1)$$
Where the last result is due to telescoping property of $T_k$. So we have a lower limit $ S_k >\color{indigo}{ 16}$
However still we cannot say $\lfloor S \rfloor = 16$ because $S$ may exceed $17$.

$$\begin{align}
\int_1^{81}\frac 1{\sqrt x}\;\;\text d x
&<\qquad\sum_{k=1}^{80}\frac 1{\sqrt k}
&&<1+\int_1^{80}\frac 1{\sqrt x}\;\;\text d x\\
\bigg[2\sqrt{x}\bigg]_1^{81}
&< \qquad\sum_{k=1}^{80}\frac 1{\sqrt k} 
&&<1+\bigg[2\sqrt{x}\bigg]_1^{80}\\
2\big(\sqrt {81}-\sqrt{1}\big)
&< \qquad\sum_{k=1}^{80}\frac 1{\sqrt k} 
&&< 1+2\big(\sqrt{80}-\sqrt{1}\big)\\
16
&< \qquad\sum_{k=1}^{80}\frac 1{\sqrt k}
&&<16.88 \end{align}$$
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NB: Wolframalpha gives $16.484$.
