Showing $\left\lfloor \sum_{k=1}^{10000} {1 \over \sqrt k}\right\rfloor = 198$ Some time ago, I was given a homework, which, among other things, asked to demonstrate the following equality:
$$\left\lfloor \sum_{k=1}^{10000} {1 \over \sqrt k}\right\rfloor = 198$$
I've tried a few things but nothing got me too far; I haven't found anything on the Internet about this, either (at least not satisfactory enough).
I've also wondered if/how it'd be possible to find the result for any positive integer $n$, i.e. to find $q$ such that 
$$\left\lfloor \sum_{k=1}^n {1 \over \sqrt k}\right\rfloor = q$$
EDIT: Thanks a lot for all your responses. This problem (at least in my case) is for 9th grade, so no integration allowed (or calculus in general), although I'm not sure a proof of that level is possible; your responses are good enough for me, anyways.
 A: In Apostol's Number Theory textbook it is shown using Euler's summation formula that for all $0<s<1$,
$$
\sum_{k=1}^n\frac{1}{k^s}=\frac{n^{1-s}-1}{1-s}+1-s\int_1^n\frac{t-\lfloor t\rfloor}{t^{s+1}}\ dt\qquad (1).
$$
Observe that for all $n>1$,
$$
0< \int_1^n\frac{t-\lfloor t\rfloor}{t^{s+1}}\ dt\leq \int_1^{\infty}\frac{1}{t^{s+1}}\ dt=\frac{1}{s}.
$$
Using these bounds in $(1)$ implies that
$$
0\leq\sum_{k=1}^n\frac{1}{k^s}-\frac{n^{1-s}-1}{1-s}<1.
$$
Consequently, it follows that
$$
\left\lfloor\sum_{k=1}^n\frac{1}{k^s}\right\rfloor=\frac{n^{1-s}-1}{1-s}
$$
whenever the right side is an integer. In particular, when $n=10^4$ and $s=\tfrac12$ the right side evaluates to the integer $198$, and this answers your question.
To answer your more general question, observe that our integer condition when $s=\tfrac12$ requires that $2(\sqrt{n}-1)$ be an integer, or equivalently, that $n$ is a perfect square. Thus,
$$
q=2(\sqrt{n}-1),
$$
whenever $n$ is a perfect square. For non-square $n$, our bounds imply that rounding $2(\sqrt{n}-1)$ will get you to within $1$ of the correct answer.
A: $\frac  1 {\sqrt {k+1}}<\int_k^{k+1}  \frac  1 {\sqrt x} dx < \frac  1 {\sqrt k}$. Sum over $k$ and use the fact that $2\sqrt x$ is an anti-derivative for $\frac 1 {\sqrt x}$ to derive the following inequality: $198-\frac 1 {100} <[\sum_{k=1}^{10000} \frac 1 {\sqrt k}]<199$. Hence the result. 
Proof of $\frac  1 {\sqrt {k+1}}<\int_k^{k+1}  \frac  1 {\sqrt x} dx < \frac  1 {\sqrt k}$: $\int_k^{k+1}  \frac  1 {\sqrt x} dx <\int_k^{k+1}  \frac  1 {\sqrt k} dx$ because $1 {\sqrt x} <1 {\sqrt k}$ when $k<x<k+1$. Hence $\int_k^{k+1}  \frac  1 {\sqrt x} dx < \frac  1 {\sqrt k}$. Similarly you get the other inequality by noting that $\frac  1 {\sqrt x}  > \frac  1 {\sqrt{ k+1}}$ when $k<x<k+1$].
A: If you draw the stair-step graph of the sum (that is, the graph is $1$ between $x=1$ and $x=2$ and it's $1/\sqrt{2}$ between $2$ and $3$, etc.)  and also draw the graph of $y=1/\sqrt{x}$, then you'll see that the sum is greater than
$$\int_1^{10001} \frac{1}{\sqrt{x}} \; dx = 198.01.$$
Then if you imagine pushing the stair-step function to the left one unit, you see that the sum is less than 
$$1+\int_1^{10000} \frac{1}{\sqrt{x}} \; dx = 199.$$
