Floor sum of reciprocal of square root of first $50$ numbers

By the same method, you find for $m>1$ $$ 2(\sqrt{51}-\sqrt m)<\sum_{k=m}^{50}\frac1{\sqrt k}<2(\sqrt{50}-\sqrt{m-1})$$ With $m=5$ (actually, $m=4$ would be enough, but is "harder to compute" as it involves more numerical square root computations), this gives us $$\begin{align}\sum_{k=1}^{50}\frac1{\sqrt k}&<1+\frac1{\sqrt 2}+\frac1{\sqrt 3}+\frac1{\sqrt 4}+2(\sqrt{50}-\sqrt 4)\\ &<1+0.71+0.58+0.5+14.15-4 \\ &=12.94\end{align}$$

$$ \sum_{k=1}^{50} \frac 1 {\sqrt{k}} = \int_1^{51} \frac 1 {\sqrt{x}} \, dx + \text{a little bit more} = 2(\sqrt{51} - 1) + \text{a little bit more}. $$ We just need to show that that "little bit" is small enough. A carefully drawn graph show show you why $$ \text{that little bit} < \left( 1 - \frac 1 {\sqrt 2} \right) + \left( \frac 1 {\sqrt 2} - \frac 1 {\sqrt 3} \right) + \left( \frac 1 {\sqrt 3} - \frac 1 {\sqrt 4} \right) + \cdots + \left( \frac 1 {\sqrt{50}} - \frac 1 {\sqrt{51}} \right) $$ and all of the terms cancel out except the first and the last, so $$ \text{that little bit} < 1 - \frac 1 {\sqrt{51}}. $$ That doesn't quite do it, so we refine the technique: $$ \frac 1 {\sqrt 1} = \int_1^2 \frac{dx}{\sqrt x} + \text{a little bit} = 2(\sqrt 2 - 1) + \text{a little bit}. $$ That little bit can be found numerically and it is less than $60\%$ of $1 - \dfrac 1 {\sqrt 2}$.

In all the latter terms, the $60\%$ would be replaced by something even smaller than $60\%$ (but always bigger than $50\%$ for reasons that should be obvious from looking at the graph).

Thus $$ \text{little bit} < 0.6\times \left( 1 - \frac 1 {\sqrt 2} \right) + \left( \frac 1 {\sqrt 2} - \frac 1 {\sqrt 3} \right) + \left( \frac 1 {\sqrt 3} - \frac 1 {\sqrt 4} \right) + \cdots + \left( \frac 1 {\sqrt{50}} - \frac 1 {\sqrt{51}} \right) $$ and that does it.