Sum based on stolz theorem The Sum is as follows:
$$ \lim_{n\to \infty}  \frac{1} {\sqrt n} \left( \frac {1} {\sqrt n}+ \frac {1} {\sqrt {n+1}} +...+ \frac{1}{\sqrt{2n}}  \right) $$.
I solved as follows using stolz theorem
$$ \lim_{n\to \infty}  \frac{x_n-x_{n-1}} {y_n-y_{n-1} } = \lim_{n\to \infty} \frac {\frac {1} {\sqrt{2n}} - \frac {1} {\sqrt{2n-1}}} {\sqrt n - \sqrt{n-1}}$$
where $$ {\{x_n\} = \left( \frac {1} {\sqrt n}+ \frac {1} {\sqrt {n+1}} +...+ \frac{1}{\sqrt{2n}}  \right)}  $$
 and
$$ {\{y_n\} = \sqrt n}  $$
I rationalized the denominator and got the answer as zero.
But the solution in the book is given as 
$$ \lim_{n\to \infty}  \frac{x_n-x_{n-1}} {y_n-y_{n-1} } = \lim_{n\to \infty} \frac {\frac {1} {\sqrt{2n}} + \frac {1} {\sqrt{2n-1}}- \frac{1} {\sqrt{n-1}}} {\sqrt n - \sqrt{n-1}}$$
$$ = 2(\sqrt2-1) $$
how they got the second step that is 
$$ \lim_{n\to \infty} \frac {\frac {1} {\sqrt{2n}} + \frac {1} {\sqrt{2n-1}}- \frac{1} {\sqrt{n-1}}} {\sqrt n - \sqrt{n-1}}$$
Please explain. 
Thanks in advance.

The second step follows from
$$
x_n - x_{n-1} = \frac{1}{\sqrt{2n}} + \frac{1}{\sqrt{2n-1}} - \frac{1}{\sqrt{n-1}},
$$
which you see by carefully noting which terms are present in $x_n$ which are not in $x_{n-1},$ and vice versa. When you go from $x_{n-1}$ to $x_n$ you gain two terms 'at the end', and loose one 'at the beginning.'
