Proof of a Simple Convergence Theorem

Let us assume that the coefficients $a_{k}$ of $S(\theta)$ are all positive and that they constitute a sequence which decreases to zero monotonically, that is, we assume the hypotheses

$$\begin{eqnarray*} a_{k} & \geq & 0,\,\forall k; \\ a_{k+1} & \leq & a_{k},\,\forall k; \\ \lim_{k\to\infty}a_{k} & =& 0. \end{eqnarray*}$$

Note however that no statement at all is made about the rate in which $a_{k}$ approaches zero. This is enough to ensure that the series is convergent for $\theta\neq 0$, as can be established by the use of the Dirichlet convergence test. Nevertheless, we will give a proof based on the geometrical representation of the series on the complex plane, because it allows us to examine the behavior of the series in the case $\theta=0$ and leads to an efficient summation technique in the convergent cases. We will show that these simple hypotheses suffice to ensure the convergence of the series $S(\theta)$ almost everywhere, in a sense which will be made clear during the argument. The result we will obtain here can be easily extended from this case to several others, but for definiteness let us consider only this case for the moment.

Interestingly, it is useful to analyze first a simple case that does not quite satisfy these hypotheses, namely the case in which $a_{k}=1$ for all values of $k$. This case does not satisfy our hypotheses since this sequence of coefficients fails to approach zero when $k\to\infty$, which implies that the series must diverge. From now on we will denote the $k=1$ basis element by $v$,

$$v = e^{\mbox{\boldmath$\imath$}\theta},$$

so that in this simple case the partial sums of the series $S(\theta)$ can be written as sums of powers of $v$,

$$S_{n}(\theta) = \sum_{k=0}^{n}e^{\mbox{\boldmath$\imath$}k\theta} = \sum_{k=0}^{n}v^{k},$$

where $v^{k}$ is a unit vector making an angle of value $k\theta$ with the real axis. In this simple case it is not difficult to verify, using the geometry of the complex plane, that the partial sums of the vectors $v^{k}$ run indefinitely around a circle, making infinite turns around it, as illustrated in Figure 2. The series fails therefore to converge, and oscillates indefinitely instead. There is, of course, an exception in the case $\theta=0$, since in this case all the vectors $v^{k}$ are equal to one and therefore the series diverges linearly to positive infinity along the real axis. For any other value of $\theta$ the partial sums turn around a circle with a certain finite radius.

Figure 2: The chain of vectors corresponding to $a_{k}=1$ for all $k$.

From the geometry shown in Figure 2 it is not difficult to verify that the radius $R$ of the circle is given by

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

The position of the center of rotation $C$ is also easily verified to be given by

$$C = \left( \frac{1}{2} , h \right) = \left( \frac{1}{2} , \frac{1}{2\tan(\theta/2)} \right),$$

so that regardless of the value of $\theta$ the center is over the vertical line $x=1/2$ of the complex plane $z=x+\mbox{\boldmath$\imath$}y$. However, it is more interesting and useful to develop a different version of the formula for the position of $C$, based solely on complex arithmetic, which produces a simple result in terms of $v$. We will do this in steps.

First, consider the first vector $v^{0}=1$ alone, and how to go from its tip to the center $C$. One can do this by going back to the middle of the vector, and then proceeding perpendicularly to it by a length $h$. The first step is accomplished by subtracting $v^{0}/2$. For the second step, we consider the unit vector $\mbox{\boldmath$\imath$}v^{0}$, which is perpendicular to $v^{0}$, being rotated from it by $\pi/2$ in the positive (counterclockwise) direction. In order to go to the center, we must then add $\mbox{\boldmath$\imath$}v^{0}h$, so that the complete path from the origin to $C$ is given arithmetically by

$$\begin{eqnarray*} C & = & v^{0} - \frac{1}{2}v^{0} + \mbox{\boldmath$\ima... ... \left( \mbox{\boldmath$\imath$}h-\frac{1}{2} \right) v^{0}. \end{eqnarray*}$$

It is easy to verify that this is the same formula for $C$ given before. Similarly, we can get to the center from the tip of the second vector $v^{1}$, and the complex arithmetic representing the complete path in this case is

$$\begin{eqnarray*} C & = & v^{0} + v^{1} - \frac{1}{2}v^{1} + \mbox{\bol... ... \left( \mbox{\boldmath$\imath$}h-\frac{1}{2} \right) v^{1}. \end{eqnarray*}$$

It is now easy to generalize this expression, representing arithmetically the path from the origin to the center, through the tip of the $n$-th vector of the chain, by

$$C = \sum_{k=0}^{n} v^{k} + \left( \mbox{\boldmath$\imath$}h-\frac{1}{2} \right) v^{n}.$$

It is clear that, since the center of the circle is a fixed point in this simple case, all these formulas, for any value of $n$, produce the same result, and that the complex number $C$ above does not really depend on $n$. The comparison of the first two formulas, for $n=0$ and $n=1$, allows us to write the factor $(\mbox{\boldmath$\imath$}h-1/2)$ in terms of $v$,

$$\begin{eqnarray*} 1 + \left( \mbox{\boldmath$\imath$}h-\frac{1}{2} \right) ... ...{\boldmath$\imath$}h-\frac{1}{2} \right) & = & \frac{v}{1-v}. \end{eqnarray*}$$

Therefore, the position of the center is given by any one of the family of formulas

$$\begin{eqnarray*} C & = & \sum_{k=0}^{n} v^{k} + \frac{v}{1-v} v^{n} \\ & = & \sum_{k=0}^{n} v^{k} + \frac{v^{n+1}}{1-v}. \end{eqnarray*}$$

We see therefore that $C$ can be written in terms of the partial sums $S_{n}$ of the $S(\theta)$ series,

$$C = S_{n} + \frac{v^{n+1}}{1-v}.$$

This is true even for the case $n=0$, which gives us the simplest form for $C$,

$$C = \frac{1}{1-v}.$$

We see here that, in the limit $\theta\to 0$, for which $v\to 1$, this point diverges to infinity, and corresponds to a circle with infinite radius that goes through the origin. This is the case in which the series diverges to infinity along the real axis, while $C$ goes to infinity along the line $x=1/2$. For any other value of $\theta$ the center is at a finite distance from the origin, and the series runs around it indefinitely.

Let us now verify what happens when we generalize this analysis to the case in which $a_{k}$ decreases monotonically to zero. We may limit the discussion to the case in which $a_{0}=1$ without loss of generality, since it is always possible to put $a_{0}$ in evidence on the whole series and rename the remaining coefficients accordingly. If we consider the new sum and corresponding chain of vectors, each vector has the same direction as before, but their lengths now decrease from $a_{0}=1$. It is easy to see that in this case the chain tends to spiral inwards within the original circle, as illustrated in Figure 3. Besides, the center of rotation is no longer fixed, but drifts from step to step in the summation sequence.

Figure 3: Example of a chain of vectors spiraling inward within the circle.

One can see that, if the coefficients $a_{k}$ approach zero very slowly, then the center also drifts very slowly, while the chain of partial sums spirals around many times. On the other hand, if the coefficients $a_{k}$ approach zero very fast, then the center drifts faster, and the chain of partial sums spirals around only a few times, if at all. The essential element of the proof of convergence is the idea of following the drift of the center, rather than the spiraling of the chain of partial sums.

We have therefore a sequence of points which we may call ``instantaneous centers of rotation'', $C_{0},C_{1},C_{2},C_{3},\ldots$, that corresponds to the sequence of partial sums $S_{0},S_{1},S_{2},S_{3},\ldots$. The same arithmetic arguments we used before can be used in this case to give $C_{n}$ in terms of $S_{n}$. We now have the partial sums

$$S_{n} = \sum_{k=0}^{n}a_{k}v^{k},$$

and the path to get to the center $C_{n}$ via the tip of the chain $S_{n}$ is represented arithmetically by

$$\begin{eqnarray*} C_{n} & = & S_{n} - \frac{1}{2}a_{n}v^{n} + \mbox{\bold... ...ft( \mbox{\boldmath$\imath$}h-\frac{1}{2} \right) a_{n}v^{n}, \end{eqnarray*}$$

where $h$ is the same as before, leading to the expression, in terms of $v$ only,

$$\begin{eqnarray*} C_{n} & = & S_{n} + \frac{v}{1-v}a_{n}v^{n} \\ & = & S_{n} + \frac{a_{n}v^{n+1}}{1-v}. \end{eqnarray*}$$

It is important to observe now that the distance from a partial sum to the corresponding center, which is given by $\vert C_{n}-S_{n}\vert$, is proportional to $a_{n}$,

$$\begin{eqnarray*} \vert C_{n}-S_{n}\vert & = & \left\vert \mbox{\boldmath$\i... ...h^{2}+1/4} \\ & = & \frac{a_{n}}{2\vert\sin(\theta/2)\vert}, \end{eqnarray*}$$

and therefore goes to zero when $a_{n}\to 0$, so long as $\theta$ is not zero or a multiple of $2\pi$. Therefore, the partial sums approach the drifting center in the $n\to\infty$ limit, and hence the two sequences, $C_{n}$ and $S_{n}$, have the same limit.

We may now consider the series $C(\theta)$ corresponding to the partial sums $C_{n}$, and the corresponding chain of complex vectors, in analogy with what we did for the $S(\theta)$ series. We will now show that, if the coefficients $a_{n}$ satisfy our hypotheses, then this series is in fact absolutely convergent, so long as $v\neq 1$. From the relation between $C_{n}$ and $S_{n}$, multiplying by $(1-v)$, we get

$$\begin{eqnarray*} (1-v)C_{n} & = & (1-v)S_{n} + a_{n}v^{n+1} \\ & = & (1-v)\sum_{k=0}^{n}a_{k}v^{k} + a_{n}v^{n+1}. \end{eqnarray*}$$

In order to write the resulting expression as a power series in $v$, me manipulate the sums and indices in order to get

$$\begin{eqnarray*} (1-v)C_{n} & = & \sum_{k=0}^{n}a_{k}v^{k} - \sum_{k=0}^{n... ...{n+1} \\ & = & a_{0} + \sum_{k=1}^{n}(a_{k}-a_{k-1})v^{k}. \end{eqnarray*}$$

If we take absolute values and use the triangular inequalities, we get

$$\begin{eqnarray*} \vert 1-v\vert\vert C_{n}\vert & \leq & \vert a_{0}\vert + \sum_{k=1}^{n}\vert a_{k}-a_{k-1}\vert, \end{eqnarray*}$$

since $v$ has unit modulus. Since the coefficients $a_{k}$ decrease monotonically to zero, we may write this as

$$\vert 1-v\vert\vert C_{n}\vert \leq a_{0} + \sum_{k=1}^{n}(a_{k-1}-a_{k}).$$

We now observe that in the remaining sum almost all terms cancel out in pairs, except for one $a_{0}$ and one $a_{n}$, so that we have

$$\vert 1-v\vert\vert C_{n}\vert \leq 2a_{0} - a_{n}.$$

If we now take the $n\to\infty$ limit, by our hypotheses $a_{n}$ vanishes, and we get

$$\vert 1-v\vert\vert C(\theta)\vert \leq 2a_{0}.$$

Therefore, since $a_{0}$ is a given finite real number, the series $(1-v)C(\theta)$ is absolutely convergent and, so long as $v\neq 1$, so is the series $C(\theta)$, for which we have

$$\vert C(\theta)\vert \leq \frac{2a_{0}}{\vert 1-v\vert}.$$

In either case the length of the chain associated to the series is finite, thus characterizing a situation of absolute convergence. This establishes the convergence of $C(\theta)$ for the case $\theta\neq 0$, and since $S_{n}$ converges to $C_{n}$, it also establishes the convergence of $S(\theta)$.

Before we give a more rigorous proof of this result, let us discuss it in terms of the two separate real series, the one with the cosines and the one with the sines. If $\theta\neq 0$, then it is clear that both series converge. The single point $\theta=0$ requires special consideration. In this case we have $v=1$, the radius of the circle becomes infinite, and the complex series diverges to infinity along the real axis. This implies that only its real part diverges, while its imaginary part is in fact identically zero. Therefore, the conclusion is that the series of sines always converges, while the series of cosines converges at all points except one, the point $\theta=0$.

We may use the geometrical ideas presented here in order to understand the origin of the discontinuity of the sine series at the special point $\theta=0$. This discontinuity will appear when $a_{k}$ decreases slowly with $k$, so that the series is not absolutely convergent, and the center of rotation drifts very little from its initial position. For $\theta$ small and positive the center $C$ will be in the upper half-plane, far from the real axis, and the chain of vectors will curve slowly upwards, away from the real axis, in order to eventually converge to that center. However, if $\theta$ is small and negative, then the center $C$ will be in the lower half-plane, again far from the real axis, and the chain of vectors will curve slowly downwards. We see therefore that there will be a discontinuity, since an infinitesimal variation that flips the sign of $\theta$ will result in a large jump in the imaginary part of the point of convergence, from strictly positive to strictly negative values.