Evaluations of Convergence

Figure 1: Illustration of the upper bounding of the sum of the terms $\vert a_{k}\vert\rho _{1}^{k}$ by an integral.

It is not a difficult task to establish the absolute and uniform convergence of a complex power series on a circle centered at the origin, or the lack thereof, starting from the behavior of the terms of the series in the limit $k\to\infty$, if we assume that they behave as inverse powers of $k$ for large values of $k$. If we have a complex power series $S_{z}$ with coefficients $a_{k}$, at a point $z_{1}$ strictly inside the unit disk,

$$S_{z} = \sum_{k=1}^{\infty}a_{k}z_{1}^{k},$$

where $\vert z_{1}\vert=\rho_{1}$ and $\rho_{1}<1$, then it is absolutely convergent if and only if the series $\overline{S}_{z}$ of the absolute values of the coefficients,

$$\begin{eqnarray*} \overline{S}_{z} & = & \sum_{k=1}^{\infty}\vert a_{k}\vert\... ...k} \\ & = & \sum_{k=1}^{\infty}\vert a_{k}\vert\rho_{1}^{k}, \end{eqnarray*}$$

converges. One can show that this sum will be finite if, for $k$ above a certain minimum value $k_{m}$, it holds that

$$\vert a_{k}\vert\rho_{1}^{k} \leq \frac{A}{k^{1+\varepsilon}},$$

for some positive real constant $A$ and some real constant $\varepsilon>0$. This is true because the sum of a finite set of initial terms is necessarily finite, and because in this case we may bound the remaining infinite sum from above by a convergent asymptotic integral,

$$\begin{eqnarray*} \sum_{k=k_{m}+1}^{\infty}\vert a_{k}\vert\rho_{1}^{k} & \leq... ...\ & = & \frac{A}{\varepsilon}\,\frac{1}{k_{m}^{\varepsilon}}, \end{eqnarray*}$$

Figure 2: Illustration of the lower bounding of the sum of the terms $\vert a_{k}\vert\rho _{1}^{k}$ by an integral.

as illustrated in Figure 1. In that illustration each vertical rectangle has base $1$ and height given by $\vert a_{k}\vert\rho _{1}^{k}$, and therefore area given by $\vert a_{k}\vert\rho _{1}^{k}$. As one can see, the construction is such that the set of all such rectangles is below the graph of the function $A/k^{1+\varepsilon}$, and therefore the sum of their areas is contained within the area under that graph, to the right of $k_{m}$. This establishes the necessary inequality between the sum and the integral.

So long as $\varepsilon$ is not zero, this establishes an upper bound to a sum of positive quantities, which is therefore a monotonically increasing sum. It then follows from the well-known theorem of real analysis that the sum necessarily converges, and therefore the series $S_{z}$ is absolutely convergent at $z_{1}$.

In addition to this, one can see that the convergence condition does not depend on $\theta$, since that dependence is only within the complex variable $z=\rho_{1}\exp(\mbox{\boldmath$\imath$}\theta)$, and vanishes when we take absolute values. This implies uniform convergence because, given a strictly positive real number $\epsilon$, absolute convergence for this value of $\epsilon$ implies convergence for this same value of $\epsilon$, with the same solution $k(\epsilon)$ for the convergence condition. This makes it clear that the solution of the convergence condition for $k$ is independent of position and therefore that the series is also uniformly convergent on the circle of radius $\rho_{1}$ centered at the origin.

This establishes a sufficient condition for the absolute and uniform convergence of the complex power series on the circle. On the other hand, if we have that, for $k$ above a certain minimum value $k_{m}$,

$$\vert a_{k}\vert\rho_{1}^{k} \geq \frac{A}{k^{1-\varepsilon}},$$

with positive real $A$ and real $\varepsilon\geq 0$, then it is possible to bound the sum $\overline{S}_{v}$ from below by an asymptotic integral that diverges to positive infinity. This is done in a way similar to the one used for the establishment of the upper bound, but inverting the situation so as to keep the area under the graph contained within the combined areas of the rectangles, as illustrated in Figure 2. The argument then establishes in this case that, for $\varepsilon>0$

$$\begin{eqnarray*} \sum_{k=k_{m}+1}^{\infty}\vert a_{k}\vert\rho_{1}^{k} & \geq... ...} + \frac{A}{\varepsilon} \lim_{k\to\infty} k^{\varepsilon}, \end{eqnarray*}$$

and therefore that $\overline{S}_{z}$ diverges to infinity. A similar calculation can be performed in the case $\varepsilon=0$, leading to logarithms and yielding the same conclusions. This does not prove or disprove convergence itself, but it does establish the absence of absolute convergence. It also shows that, so long as $\vert a_{k}\vert\rho _{1}^{k}$ behaves as a power of $k$ for large $k$, the previous condition is both sufficient and necessary for absolute convergence.