Where am I going wrong? Converging sequence I am struck with a dilemma concerning the following exercise in Bartle's Elements of Real Analysis. 

Determine the convergence or the divergence of the sequence $(x_n)$
  given by 
$$ x_n = \frac{1}{ n + 1} + \frac{1}{ n + 2}  + \cdots+ \frac{1}{ 2n}
 \; ; \;\; n \in \Bbb N $$

Genius that I am, I actually proved the sequence was not Cauchy and hence is divergent only for the hints section to tell me that $(x_n)$ is increasing and is bounded above by $ \frac{n}{n + 1} \le  1 $ and hence converges by the Monotone Convergence Theorem, which is true. This is what I did: 
I proved that $x_{n + 1} - x_n = \dfrac{ 1 }{ (2n + 2)( 2n + 1) }$ for an arbitrary natural number $n$. Then obtained a lower bound for $x_m - x_n$ for $m \gt n$ in the following manner, 
\begin{align} 
x_m - x_n& = (x_m - x_{m - 1}) + (x_{m - 1} - x_{m - 2}) + \cdots ( x_{n + 1} - x_n )\\ 
& =  \dfrac{ 1 }{ (2(m - 1) + 2)( 2(m - 1) + 1)}  + \cdots + \dfrac{ 1 }{ (2n + 2)( 2n + 1)} \\ & \gt \dfrac{ m - n }{ 2m( 2m - 1)} \gt \dfrac{ m - n }{4m} \\ 
\end{align}
Specifically if $m = 2n$ then $ x_m - x_n \gt \frac{1}{8} $. And I thought this proves the sequence is not Cauchy. This is because there is $\epsilon = \frac{1}{8}  \gt 0 \;\; $ such that given any natural number $n$ there is $m (= 2n) \gt n $ such that $|x_{m} - x_n| = x_{m} - x_n \gt \frac{1}{8}$. 
Where am I going wrong?? Any help would be gratefully appreciated. 

Your very last inequality $\frac{m-n}{2m(2m-1)}>\frac{m-n}{4m}$ is not correct. Maybe you got it by observing that $2m>2$ (which is true for $m\geq 2$) and then forgot to reverse the inequality when taking the inverse?
Actually, having proved that 
$x_{n+1}-x_n=\frac{1}{(2n+2)(2n+1)}$, you can deduce immediately that the series $\sum (x_{n+1}-x_n)$ is convergent, and hence that the sequence $(x_n)$ is convergent.
Of course, Sami's answer is much more interesting since it also gives the limit.
