If $ A=\frac{1}{2\sqrt{1}}+\frac{1}{3\sqrt{2}}+\frac{1}{4\sqrt{3}}+.........+\frac{1}{100\sqrt{99}}\;,$ Then $\lfloor A \rfloor =$ 
If $\displaystyle A=\frac{1}{2\sqrt{1}}+\frac{1}{3\sqrt{2}}+\frac{1}{4\sqrt{3}}+.........+\frac{1}{100\sqrt{99}}\;,$ Then $\lfloor A \rfloor =$
Where $\lfloor x \rfloor$ represent floor function of $x$.

$\bf{My\; Try:}$ For lower bound $$\sum^{99}_{k=1}\frac{1}{(k+1)\sqrt{k}}>\sum^{99}_{k=1}\frac{1}{(k+1)k}=\sum^{99}_{k=1}\left[\frac{1}{k}-\frac{1}{k+1}\right]=1-\frac{1}{99}$$
Now I didn't understand how can I solve it, any help?
Thanks

Let
$$A = \sum_{k=1}^{99} \frac1{(k+1) \sqrt{k}} $$
You can show that the sum is greater than $1$.  First we can lower bound the sum by
$$\sum_{k=1}^{99} \frac1{k (k+1)} = \frac{99}{100} $$
which, while less than one, may be corrected by its second term:
$$\frac{99}{100} + \frac13 \left (\frac1{\sqrt{2}} - \frac12 \right ) \gt 1$$
i.e., $A \gt 1$.  Next, consider that
$$A = \sum_{k=1}^{99} \frac1{(k+1) \sqrt{k}} = \sum_{k=1}^{99} \left (\frac{\sqrt{k}}{k} - \frac{\sqrt{k+1}}{k+1} \right ) +  \sum_{k=1}^{99} \frac{\sqrt{k+1}-\sqrt{k}}{k+1}$$
The first sum on the RHS is just $9/10$.  The second sum, however, is 
$$ \sum_{k=1}^{99} \frac1{(k+1) (\sqrt{k+1}+\sqrt{k})}  \le \frac12 \sum_{k=1}^{99} \frac1{(k+1) \sqrt{k}} = \frac12 A$$
Thus $A \le \frac9{10} + \frac12 A$, or
$$A = \sum_{k=1}^{99} \frac1{(k+1) \sqrt{k}}  \le \frac95$$
and is also greater than one.  Thus, the floor of the sum is $1$.
