How do you do this question?

We want to show the series an diverges. So we need to find a smaller function bn that diverges. That means we can eliminate options C and D, both of which converge.

an = (3 + sin(n) + e⁻ⁿ) / (2n − 1)

Sine is between -1 and +1, and e⁻ⁿ approaches 0 as n approaches infinity. So the numerator is a minimum of 2 as n approaches infinity. So we can make a new function:

an = 2 / (2n − 1)

Let's look at option B: bn = 4 / √n = 2 / (½√n).

½√n is less than 2n − 1, so bn is greater than an. We can't use that to show that an diverges.

Now let's look at option A: bn = 1 / n = 2 / (2n).

2n is greater than 2n − 1, so bn is less than an. Since bn diverges, an must also diverge.