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 $17$.

We have that $\sqrt{n+a+1}-\sqrt{n+a}$ behaves like $\frac{1}{2\sqrt{n}}$ for large values of $n$, and by picking $a=-\frac{1}{2}$ we get a telescopic term providing an accurate approximations of $\frac{1}{\sqrt{n}}$ for any $n\geq 1$:
$$ 2\sqrt{n+\tfrac{1}{2}}-2\sqrt{n-\tfrac{1}{2}} =\frac{1}{\sqrt{n}}+E(n),\qquad \frac{1}{32 n^2\sqrt{n}}\leq E(n)\leq \frac{1}{28n^2\sqrt{n}} $$
It follows that
$$ \sum_{n=1}^{80}\frac{1}{\sqrt{n}} = 2\sqrt{80+\tfrac{1}{2}}-2\sqrt{\tfrac{1}{2}}+\theta,\qquad |\theta|\leq\frac{1}{28}\zeta\left(\frac{5}{2}\right) $$
hence $\sum_{n=1}^{80}\frac{1}{\sqrt{n}}$ belongs for sure to the interval $(16,17)$ and it is pretty close to the midpoint of such interval.
