You may start by factoring out `1/(3n)` such that:

`a_n = 1/(3n)*(1/(1 + 2/(3n)) + 1/(1 + 5/(3n)) + ... + 1/(1 + (3n-1)/(3n)))`

`a_n = 1/(3n)*sum_(k=1)^n 1/(1 + (3k-1)/(3n))`

You need to notice that the sequence a_n represents the Riemann sum associated to the function `f(x) = (1/3)*(1/(x+1))` .

You need to evaluate the limit of the Riemann sum, hence, you need to evaluate the following definite integral of the function `f(x)` , such that:

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

`int_0^1 f(x)dx = int_0^1 (1/3)*(1/(x+1))dx`

`int_0^1 f(x)dx = (1/3) int_0^1(1/(x+1))dx`

`(1/3) int_0^1(1/(x+1))dx = (1/3)ln(x+1)|_0^1`

You need to use the fundamental theorem of calculus, such that:

`(1/3) int_0^1(1/(x+1))dx = (1/3)(ln 2 - ln 1)`

`(1/3) int_0^1(1/(x+1))dx = (1/3)(ln 2) => (1/3) int_0^1(1/(x+1))dx = ln (root(3)2)`

**Hence, evaluating the limit of the given sequence, using Riemann sum, yields **`lim_(n->oo) a_n = ln (root(3)2).`