Suppose we have a spiral inward towards the origin $\ f:[0,\infty)\to \mathbb{C};\ f(\theta)=r(\theta)e^{i\theta}.\ $ Does $\ \sum_n f(n)\ $ converge?

Suppose we have an inward spiral converging to the origin on an argand diagram. That is, we have a differentiable function

$$f:[0,\infty)\to \mathbb{C}\quad; f(\theta)=z = r(\theta) e^{i\theta},\quad \text{with}\quad r\overset{\theta\to\infty}{\to} 0\quad \text{and}\quad \frac{dr}{d\theta}<0\quad \forall\ \theta,$$

where $\ \theta\ $ is measured in radians, anticlockwise from an axis.

Is it true that $\ \sum_{ n\in\mathbb{N} } f(n)\ $ converges to some $\ y\in\mathbb{C}\ ?$

An obvious counter-example attempt is to make the curve spiral toward the origin very slowly, so that for each full rotation of $\ \theta\ $ our curve is always outside of a concentric circle of radius $\ \frac{1}{n}.\ $ However, the partial sums for this attempt seem to be points that lie on approximate circles and I think these circles are quite close to each other, and these circles are only getting smaller as $\ \theta\ $ increases, so I'm not sure this attempt for divergence works.

Solution 1:

Dirichlet‘s test with $a_n=r(n)$ and $b_n=e^{in}$ shows that the series is convergent under the given conditions.