Answer
$\ln$2
Work Step by Step
$\lim\limits_{x \to \infty}(1/(n+1) +1/(n+2)+...+1/2n )$
= $\lim\limits_{x \to \infty}(1/n*(1/(1+(1/n)))+1/n*(1/(1+2(1/n)))+...1/n*(1/(1+n(1/n))))$
which can be interpreted as Riemann sum with partitiony $x=1/n$
so :
$\lim\limits_{x \to \infty}(1/(n+1) +1/(n+2)+...+1/2n )$
=$ \int_{0}1/(1+x) dx = [\ln(1+x)]_{0}=\ln2$
