The integer part of $ \sum_{k=2}^{9999} \frac{1}{\sqrt{k}} $ For $k>0 $
$ \frac{1}{2\cdot \sqrt{k+1}}<\sqrt{k+1}-\sqrt{k}<\frac{1}{2\cdot \sqrt{k}} $, 
the integer part of $ \sum_{k=2}^{9999} \frac{1}{\sqrt{k}} $
My attempt here was not complying with any of the options given as my attempt was to find the min. value for the series which the question does not ask. What to do?
Options are:
A. 198
B. 197
C. 196
D. 195
I know $ \sum_{k=1}^{n} \frac{1}{k} \approx log n $ But this does not help in this case.
Also $ \sum_{k=2}^{9999} \frac{1}{\sqrt{k}} < 2(\sqrt{10000}-\sqrt{2}) $
This also does not help.

for $t\in [k,k+1]$ we have,
$$
\frac{1}{\sqrt{k+1}}\leq \frac{1}{\sqrt{t}}\leq \frac{1}{\sqrt{k}}
$$
then 
$$
\frac{1}{\sqrt{k+1}}\leq \int_k^{k+1}\frac{dt}{\sqrt{t}}\leq \frac{1}{\sqrt{k}}
$$
and we get 
\begin{eqnarray}
\sum_{k=1}^{10^4-1}\frac{1}{\sqrt{k+1}}&\leq &\sum_{k=1}^{10^4-1}\int_k^{k+1}\frac{dt}{\sqrt{t}}&\leq& \sum_{k=1}^{10^4-1}\frac{1}{\sqrt{k}}\\
\sum_{k=2}^{10^4}\frac{1}{\sqrt{k}}&\leq &\int_1^{10^4}\frac{dt}{\sqrt{t}}&\leq& \sum_{k=1}^{10^4-1}\frac{1}{\sqrt{k}}\\
\frac{1}{100}+\sum_{k=2}^{10^4-1}\frac{1}{\sqrt{k}}&\leq &\int_1^{10^4}\frac{dt}{\sqrt{t}}&\leq& 1+\sum_{k=2}^{10^4-1}\frac{1}{\sqrt{k}}\\
\frac{1}{100}+\sum_{k=2}^{10^4-1}\frac{1}{\sqrt{k}}&\leq &2(100-1)&\leq& 1+\sum_{k=2}^{10^4-1}\frac{1}{\sqrt{k}}\\
\end{eqnarray}
we can conclude that 
$$
197\leq \sum_{k=2}^{10^4-1}\frac{1}{\sqrt{k}}\leq 198-\frac{1}{100}<198
$$
so 
$$
E\left(\sum_{k=2}^{10^4-1}\frac{1}{\sqrt{k}}\right)=197
$$
