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.

You've made a mistake in evaluating $x_n-x_{n-1}$ \begin{align}x_{n}=&&\frac1{\sqrt n}+\cdots + \frac{1}{\sqrt{2n-2}}&+\frac1{\sqrt{2n-1}}+\frac1{\sqrt{2n}}\\x_{n-1}=&\frac1{\sqrt{n-1}}&+\frac1{\sqrt n}+\cdots + \frac{1}{\sqrt{2n-2}}\end{align}
They didn't.
