You may factor out `2n` at each denominator, such that:

`a_n = 1/(2n(1 + 1/(2n))) + 1/(2n(1 + 3/(2n))) + ... + 1/(2n(1 + (2n-1)/(2n))) `

You may factor out `1/(2n)` , such that:

`a_n = 1/(2n)*sum_(k=1)^n1/(1+(2k-1)/(2n))`

You need to notice that the summation that you have obtained represents a Riemann sum associated to the function `f:[0,1]->R` , f(x) = 1/(2(1+x)), having the partition` P = {[0,1/n),[1/n,2/n),[2/n,3/n),....,[(n-1)/n,1]}.`

Since the function is continuous, hence , you may evaluate the limit, such that:

`lim_(n->oo)a_n = int_0^1 f(x)dx`

Hence, you need to evaluate the definite integral of the function `f(x) = 1/(2(1+x))` , such that:

`int_0^1 f(x)dx = int_0^1 1/(2(1+x))dx`

`int_0^1 1/(2(1+x))dx = (1/2)*ln|1+x||_0^1`

Using the fundamental theorem of calculus, yields:

`int_0^1 1/(2(1+x))dx = (1/2)*(ln 2 - ln 1)`

`int_0^1 1/(2(1+x))dx = (1/2)*ln 2 = ln sqrt 2`

**Hence, evaluating the limit of the given sequence `a_n` , using Riemann sum, yields **`lim_(n->oo)a_n = ln sqrt 2.`