Value of $\left[\frac{1}{\sqrt2}+\frac{1}{\sqrt3}+\frac{1}{\sqrt4}+\cdots+\frac{1}{\sqrt{10000}}\right]$, where $[\cdot]$ is Box function How to find the value of this expression:
$$\left[\frac{1}{\sqrt2}+\frac{1}{\sqrt3}+\frac{1}{\sqrt4}+\cdots+\frac{1}{\sqrt{10000}}\right]$$
where $[\cdot]$ is Box function (greatest integer less than or equal to).
The answer is 197. I solved numerically using Wolframalpha and the answer I'm getting is correct. 
But there should must be another way of solving it, as this problem is from a 12th-grade competitive exam. I've tried to solve it using some law of surds but can't help it. Is there any analytical way of solving it?
 A: Answer Before I Noticed the Tags
Comparing to integrals, we can use that
$$
\int_1^{10001}\frac{\mathrm{d}x}{\sqrt{x}}\lt\sum_{k=1}^{10000}\frac1{\sqrt{k}}\lt1+\int_1^{10000}\frac{\mathrm{d}x}{\sqrt{x}}
$$
The left side is $2\sqrt{10001}-2=198.00999975$ (using that $\sqrt{10000}=100$, we know that the sum is greater than $198$).
The right side is $1+2\sqrt{10000}-2=199$.  
Therefore, just knowing that $\sqrt{10000}=100$, we can deduce that
$$
\left\lfloor\sum_{k=2}^{10000}\frac1{\sqrt{k}}\right\rfloor=198-1=197
$$

Algebra-Precalculus Approach
First, we have
$$
2\left(\sqrt{n+1}-\sqrt{n}\right)=\frac2{\sqrt{n+1}+\sqrt{n}}\lt\frac1{\sqrt{n}}\tag{1}
$$
Since $\sqrt{x}$ is concave, we have
$$
\frac1{\sqrt{n}}\lt\frac2{\sqrt{n+\frac12}+\sqrt{n-\frac12}}=2\left(\sqrt{n+\tfrac12}-\sqrt{n-\tfrac12}\right)\tag{2}
$$
Summing $(1)$:
$$
\begin{align}
\sum_{n=2}^{10000}\frac1{\sqrt{n}}
&\gt2\sum_{n=2}^{10000}\left(\sqrt{n+1}-\sqrt{n}\right)\\
&=2\sqrt{10001}-2\sqrt2\\
&=200.01-2.83\\
&=197.18
\end{align}
$$
Summing $(2)$:
$$
\begin{align}
\sum_{n=2}^{10000}\frac1{\sqrt{n}}
&\lt2\sum_{n=2}^{10000}\left(\sqrt{n+\tfrac12}-\sqrt{n-\tfrac12}\right)\\
&=2\sqrt{10000.5}-2\sqrt{1.5}\\
&=200.005-2.449\\
&=197.556
\end{align}
$$
