Floor sum of reciprocal of square root of first $50$ numbers 
Find Sum of $$\bigg\lfloor 1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{4}}+\cdots+\frac{1}{\sqrt{50}}\bigg\rfloor$$

$\bf{My\; Try::}$ Let $\displaystyle y=f(x) = \frac{1}{\sqrt{x}}\;,$ Then draw that graph in coordinate axis, We get
$$\displaystyle \int_{1}^{51}\frac{1}{\sqrt{x}}dx<\sum^{50}_{k=1}\frac{1}{\sqrt{k}}<1+\int_{1}^{50}\frac{1}{\sqrt{x}}dx$$
So we get $$\displaystyle 2\left(\sqrt{51}-1\right)<\sum^{50}_{k=1}\frac{1}{\sqrt{k}}<1+2(\sqrt{50}-1)$$
So we get $$12.28<\sum^{50}_{k=1}\frac{1}{\sqrt{k}}<13.14$$
But answer given is $12,$ How cai i solve avobe question, Help Required, Thanks

$$
\sum_{k=1}^{50} \frac 1 {\sqrt{k}} = \int_1^{51} \frac 1 {\sqrt{x}} \, dx + \text{a little bit more} = 2(\sqrt{51} - 1) + \text{a little bit more}.
$$
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.
