Answer
We show that
$\mathop {\lim }\limits_{n \to + \infty } {a_n} = \ln 2$
Work Step by Step
We have the general term:
${a_n} = \dfrac{1}{n}\mathop \sum \limits_{k = 1}^n \dfrac{1}{{1 + \left( {k/n} \right)}}$
Let us consider a closed interval $\left[ {0,1} \right]$. We partition the interval into $n$ subintervals such that each subinterval have the same length given by $\Delta x = \dfrac{{1 - 0}}{n} = \dfrac{1}{n}$. A middle point in each subinterval is defined by ${x_k} = \dfrac{k}{n}$, for $k = 1,2,...n - 1$.
Write $f\left( {{x_k}} \right) = \dfrac{1}{{1 + \left( {k/n} \right)}} = \dfrac{1}{{1 + {x_k}}}$.
Thus, the general term of the sequence can be written as
${a_n} = \dfrac{1}{n}\mathop \sum \limits_{k = 1}^n \dfrac{1}{{1 + \left( {k/n} \right)}} = \mathop \sum \limits_{k = 1}^n f\left( {{x_k}} \right)\Delta x$
Next, we evaluate the limit:
$\mathop {\lim }\limits_{n \to + \infty } {a_n} = \mathop {\lim }\limits_{n \to + \infty } \mathop \sum \limits_{k = 1}^n f\left( {{x_k}} \right)\Delta x$
Since $\Delta x = \dfrac{1}{n}$. The limit above is equivalent to
$\mathop {\lim }\limits_{n \to + \infty } {a_n} = \mathop {\lim }\limits_{\Delta x \to 0} \mathop \sum \limits_{k = 1}^n f\left( {{x_k}} \right)\Delta x$
By definition 4.5.1 of Section 4.5, the limit on the right-hand side is the Riemann sum of a definite integral:
$\mathop \smallint \limits_0^1 f\left( x \right){\rm{d}}x$
Thus,
$\mathop {\lim }\limits_{n \to + \infty } {a_n} = \mathop {\lim }\limits_{\Delta x \to 0} \mathop \sum \limits_{k = 1}^n f\left( {{x_k}} \right)\Delta x = \mathop \smallint \limits_0^1 f\left( x \right){\rm{d}}x$
$\mathop {\lim }\limits_{n \to + \infty } {a_n} = \mathop \smallint \limits_0^1 \dfrac{1}{{1 + x}}{\rm{d}}x = \left[ {\ln \left( {1 + x} \right)} \right]_0^1 = \ln 2$
Hence, $\mathop {\lim }\limits_{n \to + \infty } {a_n} = \ln 2$.
