Test the following series for convergence or divergence. Justify the decision.

First, for , we have

(We know this since is decreasing on the interval , and so the area of the rectangle of width and height is greater than the area under the curve .) Then, this implies

But, we know converges; hence,

also converges.

If f(x) is decreasing, shouldn’t the area of the rectangle[with width 1/n and height f(1/n)] be less than the integral..?

since, the interval on which the rectangle stands is (0,1/n)

Yep, I think this answer is incorrect. This can be solved with comparison test, where the upper bound can be computed with the monotonicity of integrals as

it’s increasing on that interval