TITLE: Finding the Integer part of $\sum_{k=2} ^{9999}\frac{1}{\sqrt k}$
QUESTION [5 upvotes]: Question from Model Question Paper for B.Math/B.Stat: Page 28, Question 27 by Indian Statistical Institute
Q. Assume the following inequalities for positive k:
$$\frac{1}{2\sqrt{k+1}}< \sqrt{k+1}-\sqrt k<\frac{1}{2\sqrt k}$$
The Integer part of: 
$$\sum_{k=2} ^{9999}\frac{1}{\sqrt k}$$ equals:
A. $198$
B. $197$
C. $196$
D. $195$
My approach:
$$S=\sum_{k=2} ^{9999}\frac{1}{\sqrt k}$$
$$S=\frac{1}{\sqrt 2}+\frac{1}{\sqrt 3}+ \cdots + \frac{1}{\sqrt{9999}}$$
$$S=2\left(\frac{1}{2\sqrt 2}+\frac{1}{2\sqrt 3}+ \cdots + \frac{1}{2\sqrt{9999}}\right)$$
$$S\approx 2\left((\sqrt 2- \sqrt 1) + (\sqrt3- \sqrt2)+\cdots +(\sqrt{9999}-\sqrt{9998})\right)$$
$$[S]\approx2(\sqrt{9999}-\sqrt1)$$
$$[S]\approx 2(98)$$
$$[S]=196$$
I know my method is not an efficient one.
Also I don't know if the answer is correct. Any help will be appreciated.

REPLY [6 votes]: $\sum\limits_{k=1}^{9999}\dfrac{1}{\sqrt{k+1}}<2(\sum\limits_{k=1}^{9999}{\sqrt{k+1}}-\sqrt k)< \sum\limits_{k=1}^{9999}\dfrac{1}{\sqrt{k}}$.
Now, the sum $2(\sum\limits_{k=1}^{9999}{\sqrt{k+1}}-\sqrt k)$ telescopes, to give us 198.
We thus have:
$\sum\limits_{k=1}^{9999}\dfrac{1}{\sqrt{k+1}}< 198$. So, $\sum\limits_{k=2}^{10,000}\dfrac{1}{\sqrt{k}} < 198$, and
$\sum\limits_{k=1}^{9999}\dfrac{1}{\sqrt{k}} > 198$.
Now do the obvious little manipulations, to get your answer.

REPLY [2 votes]: Hint: If we analyze what you have done, the procedure you used yields the fact that $S\lt 2(\sqrt{9999}-\sqrt{1})\approx 197.99$.
You need to use the other inequality to show in the same way that $S\gt 2(\sqrt{10000}-\sqrt{2})$.
The two results combined will give you the answer.