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

This series is a Riemann sum for $$\int_1^{99} \frac {dx}{(1+x)\sqrt x}$$
But that integral can be computed "exactly", as the indefinite integral is $2\,tan^{-1}(\sqrt x)$.  We get about $1.37$ for the definite integral, and it's not hard to see that this makes your answer $1$.
