About the Ratio Test

Let us show that the Taylor series of an analytic function cannot satisfy the ratio test at the boundary of its maximum convergence disk. We will assume that the convergence disk is the open unit disk, but the proof can be easily generalized. It the series satisfies the test, then it follows that there is a strictly positive real number $q<1$ and an integer $k_{m}$ such that, for all $k>k_{m}$ we have

$$\frac{\vert a_{k+1}\vert}{\vert a_{k}\vert}\,\vert z\vert < q.$$

We observe now that since $\vert z\vert=1$ over the whole unit circle, this condition is in fact independent of the angular position and valid over the whole unit circle, so that we have

$$\frac{\vert a_{k+1}\vert}{\vert a_{k}\vert} < q.$$

Now, consider a point $z_{1}$ given by any value of $\theta$ and the value

$$\vert z_{1}\vert = \frac{q+1}{2q}$$

for the radius. Since $q$ is strictly positive and strictly smaller than one, we have that

$$\begin{eqnarray*} q & < & 1 \;\;\;\Rightarrow \\ q+q & < & q+1 \;\;\;\Rightarrow \\ \frac{q+1}{2q} & > & 1. \end{eqnarray*}$$

It follows therefore that the point $z_{1}$ is strictly outside the unit disk. We consider now the ratio test for this new point. We start from the known valid condition on the unit disk, so that we have

$$\begin{eqnarray*} \frac{\vert a_{k+1}\vert}{\vert a_{k}\vert} & < & q \;\;\;... ...a_{k+1}\vert}{\vert a_{k}\vert}\,\vert z_{1}\vert & < & q_{1}, \end{eqnarray*}$$

which establishes the upper bound $q_{1}$. Besides, since we know that $q$ is strictly less than one, we also have that

$$\begin{eqnarray*} q & < & 1 \;\;\;\Rightarrow \\ q+1 & < & 1+1 \;\;\;\Rightarrow \\ \frac{q+1}{2} & < & 1. \end{eqnarray*}$$

What we have concluded here is that $q_{1}<1$, and therefore we conclude that the series satisfies the ratio test at the new point, which is strictly outside the unit disk, and hence converges there. Therefore, the basic convergence theorem implies that the maximum disk of convergence of the series $S_{z}$ extends beyond the unit circle. This contradicts the hypothesis that the open unit disk is the maximum disk of convergence, and we conclude therefore that the series cannot satisfy the ratio test at the boundary of its maximum disk of convergence.

Another way to interpret this result is to say that, if the Taylor series $S_{z}$ does satisfy the ratio test at the boundary of a given disk, then that disk is not its maximum disk of convergence. It follows therefore that the maximum disk of convergence is larger than the given disk, and contains it.