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
 A: We have:
$$ \frac{1}{\sqrt{k}}-\frac{1}{\sqrt{k+1}} = \frac{\sqrt{k+1}-\sqrt{k}}{\sqrt{k(k+1)}}=\frac{1}{k\sqrt{k+1}+(k+1)\sqrt{k}}\geq \frac{1}{2(k+1)\sqrt{k}}$$
hence:
$$ \sum_{k=1}^{99}\frac{1}{(k+1)\sqrt{k}}\leq 2\sum_{k=1}^{99}\left(\frac{1}{\sqrt{k}}-\frac{1}{\sqrt{k+1}}\right)=2-\frac{1}{5}$$
but in a similar way:
$$ \sum_{k=1}^{99}\frac{1}{(k+1)\sqrt{k}}\geq \frac{1}{2}+2\sum_{k=2}^{99}\left(\frac{1}{\sqrt{k-1/4}}-\frac{1}{\sqrt{k+3/4}}\right)=\frac{1}{2}+\frac{4}{\sqrt{7}}-\frac{4}{\sqrt{399}}$$
so $\lfloor A \rfloor = \color{red}{1}$. Another great victory for creative telescoping.
A: 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 $1$.
A: 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 $1$.
A: Hint. Observe that, for  $x\in [k,k+1],\, k=1,2,3,\cdots$,
$$
\frac1{(k+2)\sqrt{k+1}} \leq\frac1{(x+1)\sqrt{x}}\leq \frac1{(k+1)\sqrt{k}}.
$$
Integrating both sides from $x=k$ to $x=k+1$ gives
$$
\frac1{(k+2)\sqrt{k+1}} \leq\int_k^{k+1}\frac1{(x+1)\sqrt{x}}dx\leq \frac1{(k+1)\sqrt{k}}
$$ then, summing from $k=1$ to $k=99$,
$$
\sum_{k=1}^{100}\frac1{(k+1)\sqrt{k}}-\frac12\leq\int_1^{100}\frac1{(x+1)\sqrt{x}}dx\leq \sum_{k=1}^{99}\frac1{(k+1)\sqrt{k}}.
$$ We have
$$
\int_1^{100}\frac1{(x+1)\sqrt{x}}dx=2\arctan (10) -\frac{\pi}2.
$$
It follows that

$$
2\arctan (10) -\frac{\pi}2\leq\sum_{k=1}^{99}\frac1{(k+1)\sqrt{k}}\leq2\arctan (10) -\frac{\pi}2+\frac12-\frac1{1010}.
$$ or
  $$
\color{red}{1.371\cdots}<\sum_{k=1}^{99}\frac1{(k+1)\sqrt{k}}<\color{blue}{1.870\cdots}.
$$

