Final answer:
The given series is divergent according to the integral test.
Step-by-step explanation:
To determine whether the series ∑(n = 1 to ∞) 1/n√3 is convergent or divergent, we can use the integral test. The integral test states that if the integral of the function from 1 to infinity converges, then the series also converges, and if the integral diverges, then the series also diverges.
In this case, we need to evaluate the integral of 1/x√3 from 1 to infinity. Let's set u = x√3, so du = √3dx. This transforms the integral into ∫(1/u) * √3 * du = √3∫(1/u) du.
The integral of 1/u is ln|u|, so we have √3 ln|u| evaluated from 1 to infinity. As u = x√3, when x approaches infinity, u also approaches infinity. Thus, √3 ln|u| evaluated from 1 to infinity is equal to √3 ln|∞| - √3 ln|1|. Since ln|∞| is infinity and ln|1| is 0, the integral is equal to infinity. Therefore, the series is divergent.