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 $17$.
 A: $$\begin{align}
\int_1^{81}\frac 1{\sqrt x}\;\;\text d x
&<\qquad\sum_{k=1}^{80}\frac 1{\sqrt k}
&&<1+\int_1^{80}\frac 1{\sqrt x}\;\;\text d x\\
\bigg[2\sqrt{x}\bigg]_1^{81}
&< \qquad\sum_{k=1}^{80}\frac 1{\sqrt k} 
&&<1+\bigg[2\sqrt{x}\bigg]_1^{80}\\
2\big(\sqrt {81}-\sqrt{1}\big)
&< \qquad\sum_{k=1}^{80}\frac 1{\sqrt k} 
&&< 1+2\big(\sqrt{80}-\sqrt{1}\big)\\
16
&< \qquad\sum_{k=1}^{80}\frac 1{\sqrt k}
&&<16.88 \end{align}$$
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NB: Wolframalpha gives $16.484$.
A: 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.
A: You can use Abel Sum: 
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 
$$\left\lfloor S\left(80\right)\right\rfloor =16.$$
A: If you now use the fact that $\tfrac{a_k}2<T_{k-1}$ (because $\frac1{2\sqrt k}<\frac1{\sqrt k +\sqrt{k-1}}=T_{k-1}$) you'll find that the sum has to be strictly smaller than $2\sqrt{80}$, so you now know that the answer is either $16$ or $17$.
But a little more detail shows that
$$\sum_{k=2}^{80}\frac1{\sqrt{k}}<2\sum_{k=2}^{80}\left(\sqrt k - \sqrt{k-1}\right)=2(\sqrt{80}-1)<16;$$
but then
$$\sum_{k=1}^{80}\frac1{\sqrt{k}}=\frac1{\sqrt 1}+\sum_{k=2}^{80}\frac1{\sqrt{k}}<1+16=17$$
(note that this is strict; actually $2(\sqrt{80}-1)+1\simeq16.889$). So the result is $16$.
