A Monotonicity Test

Within the set of all series $S_{v}$ which are very weakly convergent, there is a subset of series that all converge almost everywhere, as we will now show. This is the class of series $S_{v}$ that have coefficients $a_{k}$ that converge monotonically to zero. This monotonicity of the coefficients can, therefore, be used as a convergence test. No statement at all has to be made about the speed of convergence to zero, but we may as well focus our attention on series of type $S_{v,h0}$, with coefficients behaving as $\vert a_{k}\vert\propto 1/k^{p}$ with $0<p\leq 1$, which are those for which this analysis is most useful, since in this case the sum

$$\sum_{k=1}^{\infty} \vert a_{k}\vert$$

diverges to infinity and the series is therefore not absolutely or uniformly convergent. For simplicity we will take the case in which $a_{k}>0$ for all $k$, but the argument can be easily generalized to several other cases, as will be discussed later on. Let us consider then the series

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

where $a_{k}>0$ for all $k$, $a_{k+1}\leq a_{k}$ for all $k$ and

$$\lim_{k\to\infty}a_{k} = 0.$$

Such a series can be shown to converge for $\theta\neq 0$ by the use of the Dirichlet test. For $\theta=0$ it diverges to positive infinity along the real axis, and therefore satisfies the hypotheses defining a series of type $S_{v,h0}$ in a regular integral-differential chain. It is not too difficult to show that, for any non-zero value of $\theta$ in the periodic interval, the Dirichlet partial sums

$$D_{N} = \sum_{k=1}^{N} v^{k},$$

are contained within a closed disk of radius

$$R = \frac{1}{2\vert\sin(\theta/2)\vert},$$

centered at the point

$$C = \frac{1}{2} + \mbox{\boldmath$\imath$}\, \frac{1}{2\tan(\theta/2)}$$

of the complex plane, for all values of $N$. The absolute values $\vert D_{N}\vert$ of the Dirichlet partial sums are therefore bounded by a constant for all $N$, and since the coefficients go monotonically to zero, the Dirichlet test applies and the series $S_{v}$ is convergent.

We will, however, demonstrate the convergence in another way, which we believe to be more fruitful, and more directly related to our problem here. What we will do is to construct another expression, involving another DP Fourier series, that converges to the same function and, unlike $S_{v}$, does so absolutely and uniformly almost everywhere. Consider then the following algebraic passage-work for our series, for $\theta\neq 0$, which is equivalent to $v\neq 1$. We start by multiplying and dividing the series by the factor $(v-1)$, and distributing the one in the numerator,

$$\begin{eqnarray*} S_{v} & = & \frac{v-1}{v-1} \sum_{k=1}^{\infty} a_{k}v^{k... ...fty} a_{k}v^{k+1} - \sum_{k=1}^{\infty} a_{k}v^{k} \right]. \end{eqnarray*}$$

We now change the index of the first series, in order to be able to join the two resulting series,

$$\begin{eqnarray*} S_{v} & = & \frac{1}{v-1} \left[ \, \sum_{k=2}^{\infty} ... ... a_{1}v + \sum_{k=2}^{\infty} (a_{k-1}-a_{k})v^{k} \right]. \end{eqnarray*}$$

If we now define the new coefficients $b_{1}=-a_{1}$ and $b_{k}=a_{k-1}-a_{k}$ for $k>1$, we have a new series $C_{v}$, which we name the center series of $S_{v}$, so that we have

$$\begin{eqnarray*} S_{v} & = & \frac{1}{v-1}\, C_{v}, \\ C_{v} & = & \sum_{k=1}^{\infty} b_{k}v^{k}. \end{eqnarray*}$$

The name we chose for this series comes from the fact that it describes the relatively small drift in the complex plane of the instantaneous center of rotation of the convergence process of the $S_{v}$ series. Specially for $a_{k}\propto 1/k^{p}$ with the values of $p$ closer to zero, the convergence process of the $S_{v}$ series proceeds in a long, slow spiral around the limiting point, while that of the $C_{v}$ series goes more or less directly to it. Let us now show that $C_{v}$ is absolutely and thus uniformly convergent. We simply consider the series $\overline{C}_{v}$ of the absolute values of the terms of $C_{v}$. Since $\vert v\vert=1$, we get

$$\begin{eqnarray*} \overline{C}_{v} & = & \sum_{k=1}^{\infty} \vert b_{k}\ver... ... & = & a_{1} + \sum_{k=2}^{\infty} \vert a_{k-1}-a_{k}\vert. \end{eqnarray*}$$

Since the coefficients $a_{k}$ decrease monotonically to zero, we have $\vert a_{k-1}-a_{k}\vert=a_{k-1}-a_{k}$ and thus we may write

$$\begin{eqnarray*} \overline{C}_{v} & = & a_{1} + \sum_{k=2}^{\infty} a_{k-... ...nfty} a_{k} - \sum_{k=2}^{\infty} a_{k} \\ & = & 2a_{1}. \end{eqnarray*}$$

Since $a_{1}$ is some finite real number, this establishes an upper bound for $\overline{C}_{v}$. Since this series is a sum of positive terms and thus monotonically increasing, this implies that it converges and therefore that $C_{v}$ is absolutely convergent. Since the bound and therefore the criterion of convergence are independent of $\theta$, the convergence is also uniform. Therefore, for $v\neq 1$ we may evaluate $S_{v}$ by first evaluating $C_{v}$ and then multiplying the result by the simple pole $1/(v-1)$.

This proves not only the absolute and uniform convergence of the series $C_{v}$ everywhere on the unit circle, it also implies the convergence of the original series $S_{v}$ at all points except the special point $v=1$, which corresponds to $\theta=0$. Obviously the $S_{v}$ series does not converge absolutely or uniformly, but it does converge at all points except $v=1$. We see therefore that the monotonicity of the coefficients can be used as a test for simple point-wise convergence of the $S_{v}$ series almost everywhere on the unit circle. The special point, were the $S_{v}$ series diverges, is easily identified as the point where $w(z)$ has a borderline hard singularity, since a single logarithmic integration of $w(z)$ will necessarily result in a soft singularity there.

If we extend the series $C_{v}$ to a full complex power series $C_{z}$ on the unit disk, just as we did before in the case of $S_{v}$ and $S_{z}$, we immediately see that

$$C_{z} = (z-1)S_{z}.$$

Note that $C_{z}$ satisfies all the necessary conditions for convergence to an inner analytic function. Since $S_{z}$ converges to an analytic function on the open unit disk and $(z-1)$ is analytic on the whole complex plane, it follows that $C_{z}$ also converges to an analytic function on that disk. In addition to this, since $(z-1)$ reduces to a real function on the real line and $S_{z}$ reduces to a real function on the real interval $(-1,1)$, so does $C_{z}$. Finally, it is clear that since $S_{z}$ is zero at $z=0$, so is $C_{z}$. Therefore $C_{z}$ converges to an inner analytic function $\gamma(z)$, so that we have

$$\gamma(z) = (z-1)w(z).$$

Since $C_{z}$ is absolutely and uniformly convergent on the unit circle, the function $\gamma(z)$ can have only soft singularities on that circle. However, $S_{z}$ diverges to infinity at one point on the unit circle and, due to the extended version of Abel's theorem [2], it follows that $w(z)$ must have a hard singularity at that point. We see therefore that in this case the multiplication by $(z-1)$ has the same effect of a logarithmic integration. While the function $w(z)$ has a single borderline hard singularity at $z=1$, the function $\gamma(z)$ has a borderline soft singularity at that point.

We may conclude here that this whole class of series satisfies our hypotheses defining a series of type $S_{v,h0}$ in a regular integral-differential chain, as well as that all of them converge almost everywhere. Besides, all the series in this whole class have a single dominant borderline hard singularity located at $z=1$. Note that even if a monotonic series has coefficients that go to zero as a power in ways other than $1/k^{p}$ with $0<p\leq 1$, the construction of $C_{z}$ still applies. In this case $w(z)$ will still have a dominant singularity at $z=1$, although no longer a borderline hard one, but a soft one instead.