Rigorous Proof of the Theorem

The same complex arithmetic used in the previous section can be employed for the construction a rigorous and direct demonstration that the sequence $S_{n}$ is convergent. The argument is very similar to a demonstration of the Dirichlet convergence test. In order to do this we show below that, for $v\neq 1$, the sequence $S_{n}$ is a Cauchy sequence within a closed disc in the complex plane, which implies that it is convergent. The demonstration is quite simple and short. We simply consider the partial sums

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

multiply by $(1-v)$ and manipulate the summation indices in order to obtain, in a way similar to the manipulations done before,

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

Taking now absolute values and using the triangular inequalities we get

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

Assuming now that the sequence of coefficients $a_{k}$ decreases monotonically to zero, we may write

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

Once again almost all terms in the remaining sum cancel out in pairs, and we are left with

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

It follows that, so long as $v\neq 1$, we may write a finite upper bound to the partial sum,

$$\vert S_{n}\vert \leq \frac{2a_{0}}{\vert 1-v\vert} = \rho,$$

which does not depend on $n$, and is therefore valid for all $n$. We conclude therefore that the whole set of partial sums is contained within a closed disc of radius $\rho$, centered at the origin, which is the initial point of the sequence. As a consequence, given any two partial sums $S_{n}$ and $S_{n'}$, the distance between them is less than $2\rho$,

$$\vert S_{n}-S_{n'}\vert \leq 2\rho,$$

for any $n$ and any $n'$. It is not difficult to repeat this argument for a sum that starts at an arbitrary intermediate point $m$ of the sequence, and goes up to a point $n>m$. This is the difference of two partial sums of the series,

$$\begin{eqnarray*} S_{mn} & = & S_{n}-S_{m} \\ & = & \sum_{k=m+1}^{n}a_{k}v^{k}. \end{eqnarray*}$$

Repeating for $S_{mn}$ the same manipulations executed before for $S_{n}$, we get this time

$$(1-v)S_{mn} = a_{m+1}v^{m+1} + \sum_{k=m+2}^{n}(a_{k}-a_{k-1})v^{k} - a_{n}v^{n+1}.$$

Taking now absolute values we get

$$\vert 1-v\vert\vert S_{mn}\vert \leq 2a_{m+1}.$$

In this way we see that, so long as $v\neq 1$, all the elements of the sequence $S_{n}$ with $n>m$ are within a closed disc centered at $S_{m}$, with a finite radius $\rho_{m+1}$,

$$\vert S_{mn}\vert = \vert S_{n}-S_{m}\vert \leq \frac{2a_{m+1}}{\vert 1-v\vert} = \rho_{m+1},$$

which is independent of $n$, so that this relation is valid for all $n$. It follows that any two elements $S_{n}$ and $S_{n'}$ of the sequence such that $n>m$ and $n'>m$ are within this disc, and hence the distance between them is bounded by the diameter of the disc,

$$\vert S_{n'}-S_{n}\vert \leq 2\rho_{m+1}.$$

This establishes the Cauchy-sequence structure of our sequence $S_{n}$. In order to complete the argument, let a real number $\epsilon>0$ be given. We consider the infinite collection of positive real numbers

$$2\rho_{m+1} = \frac{4a_{m+1}}{\vert 1-v\vert},$$

all of which are finite so long as $v\neq 1$. Since we have by our hypotheses that $a_{m+1}\to 0$ when $m\to\infty$, it follows that there is a value of $m$ such that

$$2\rho_{m+1} = \frac{4a_{m+1}}{\vert 1-v\vert} < \epsilon.$$

If we consider this value of $m$, then it follows that, for any $n>m$ and any $n'>m$, it is true that

$$\vert S_{n'}-S_{n}\vert \leq 2\rho_{m+1} = \frac{4a_{m+1}}{\vert 1-v\vert} < \epsilon,$$

which establishes that there is a value of $m$ that satisfies the criterion for a Cauchy sequence. Since $S_{n}$ is thus shown to be a Cauchy sequence within a closed disc, which is a complete set, it follows that it converges. As a consequence, so long as $v\neq 1$ the Fourier cosine and sine series that correspond to the sequence $S_{n}$ are both convergent. The same comments made before about the special case $v=1$ still apply, of course, so that the sine series converges everywhere, while the cosine series converges at all points except $\theta=0$.