191. Power Series.

One of the most important parts of the theory of the ordinary functions which occur in elementary analysis (such as the sine and cosine, and the logarithm and exponential, which will be discussed in the next chapter) is that which is concerned with their expansion in series of the form \(\sum a_{n}x^{n}\). Such a series is called a power series in \(x\). We have already come across some cases of expansion in series of this kind in connection with Taylor’s and Maclaurin’s series (§ 148). There, however, we were concerned only with a real variable \(x\). We shall now consider a few general properties of power series in \(z\), where \(z\) is a complex variable.

A. A power series \(\sum a_{n}z^{n}\) may be convergent for all values of \(z\), for a certain region of values, or for no values except \(z = 0\).

It is sufficient to give an example of each possibility.

1. The series \(\sum \dfrac{z^{n}}{n!}\) is convergent for all values of \({z}\). For if \(u_{n} = \dfrac{z^{n}}{n!}\) then \[|u_{n+1}|/|u_{n}| = |z|/(n + 1) \to 0\] as \(n \to \infty\), whatever value \(z\) may have. Hence, by d’Alembert’s Test, \(\sum |u_{n}|\) is convergent for all values of \(z\), and the original series is absolutely convergent for all values of \(z\). We shall see later on that a power series, when convergent, is generally absolutely convergent.

2. The series \(\sum n!\, z^{n}\) is not convergent for any value of \(z\) except \(z = 0\). For if \(u_{n} = n!\, z^{n}\) then \(|u_{n+1}|/|u_{n}| = (n + 1)|z|\), which tends to \(\infty\) with \(n\), unless \(z = 0\). Hence (cf. Ex. XXVII. 1, 2, 5) the modulus of the \(n\)th term tends to \(\infty\) with \(n\); and so the series cannot converge, except when \(z = 0\). It is obvious that any power series converges when \(z = 0\).

3. The series \(\sum z^{n}\) is always convergent when \(|z| < 1\), and never convergent when \(|z| \geq 1\). This was proved in § 88. Thus we have an actual example of each of the three possibilities.



B. If a power series \(\sum a_{n}z^{n}\) is convergent for a particular value of \(z\), say \(z_{1} = r_{1}(\cos\theta_{1} + i\sin\theta_{1})\), then it is absolutely convergent for all values of \(z\) such that \(|z| < r_{1}\).

For \(\lim a_{n}z_{1}^{n} = 0\), since \(\sum a_{n}z_{1}^{n}\) is convergent, and therefore we can certainly find a constant \(K\) such that \(|a_{n}z_{1}^{n}| < K\) for all values of \(n\). But, if \(|z| = r < r_{1}\), we have \[|a_{n}z^{n}| = |a_{n}z_{1}^{n}| \left(\frac{r}{r_{1}}\right)^{n} < K \left(\frac{r}{r_{1}}\right)^{n},\] and the result follows at once by comparison with the convergent geometrical series \(\sum (r/r_{1})^{n}\).

In other words, if the series converges at \(P\) then it converges absolutely at all points nearer to the origin than \(P\).

Example. Show that the result is true even if the series oscillates finitely when \(z = z_{1}\). [If \(s_{n} = a_{0} + a_{1}z_{1} + \dots + a_{n}z_{1}^{n}\) then we can find \(K\) so that \(|s_{n}| < K\) for all values of \(n\). But \(|a_{n}z_{1}^{n}| = |s_{n} – s_{n-1}| \leq |s_{n-1}| + |s_{n}| < 2K\), and the argument can be completed as before.]


193. The region of convergence of a power series. The circle of convergence.

Let \(z = r\) be any point on the positive real axis. If the power series converges when \(z = r\) then it converges absolutely at all points inside the circle \(|z| = r\). In particular it converges for all real values of \(z\) less than \(r\).

Now let us divide the points \(r\) of the positive real axis into two classes, the class at which the series converges and the class at which it does not. The first class must contain at least the one point \(z = 0\). The second class, on the other hand, need not exist, as the series may converge for all values of \(z\). Suppose however that it does exist, and that the first class of points does include points besides \(z = 0\). Then it is clear that every point of the first class lies to the left of every point of the second class. Hence there is a point, say the point \(z = R\), which divides the two classes, and may itself belong to either one or the other. Then the series is absolutely convergent at all points inside the circle \(|z| = R\).

For let \(P\) be any such point. We can draw a circle, whose centre is \(O\) and whose radius is less than \(R\), so as to include \(P\) inside it. Let this circle cut \(OA\) in \(Q\). Then the series is convergent at \(Q\), and therefore, by Theorem B, absolutely convergent at \(P\).

On the other hand the series cannot converge at any point \(P’\) outside the circle. For if it converged at \(P’\) it would converge absolutely at all points nearer to \(O\) than \(P\); and this is absurd, as it does not converge at any point between \(A\) and \(Q’\) (Fig. 51).

So far we have excepted the cases in which the power series (1) does not converge at any point on the positive real axis except \(z = 0\) or (2) converges at all points on the positive real axis. It is clear that in case (1) the power series converges nowhere except when \(z = 0\), and that in case (2) it is absolutely convergent everywhere. Thus we obtain the following result:

a power series either

(1) converges for \(z = 0\) and for no other value of \(z\); or

(2) converges absolutely for all values of \(z\); or

(3) converges absolutely for all values of \(z\) within a certain circle of radius \(R\), and does not converge for any value of \(z\) outside this circle.

In case (3) the circle is called the circle of convergence and its radius the radius of convergence of the power series.

It should be observed that this general result gives absolutely no information about the behaviour of the series on the circle of convergence. The examples which follow show that as a matter of fact there are very diverse possibilities as to this.

Example LXXX
1. The series \(1 + az + a^{2}z^{2} + \dots\), where \(a > 0\), has a radius of convergence equal to \(1/a\). It does not converge anywhere on its circle of convergence, diverging when \(z = 1/a\) and oscillating finitely at all other points on the circle.

2. The series \(\dfrac{z}{1^{2}} + \dfrac{z^{2}}{2^{2}} + \dfrac{z^{3}}{3^{2}} + \dots\) has its radius of convergence equal to \(1\); it converges absolutely at all points on its circle of convergence.

3. More generally, if \(|a_{n+1}|/|a_{n}| \to \lambda\), or \(|a_{n}|^{1/n} \to \lambda\), as \(n \to \infty\), then the series \(a_{0} + a_{1}z + a_{2}z^{2} + \dots\) has \(1/\lambda\) as its radius of convergence. In the first case \[\lim |a_{n+1}z^{n+1}|/|a_{n}z^{n}| = \lambda |z|,\] which is less or greater than unity according as \(|z|\) is less or greater than \(1/\lambda\), so that we can use d’Alembert’s Test (§ 168, 3). In the second case we can use Cauchy’s Test (§ 168, 2) similarly.

4. The logarithmic series. The series \[z – \tfrac{1}{2} z^{2} + \tfrac{1}{3} z^{3} – \dots\] is called (for reasons which will appear later) the ‘logarithmic’ series. It follows from Ex. 3 that its radius of convergence is unity.

When \(z\) is on the circle of convergence we may write \(z = \cos\theta + i\sin\theta\), and the series assumes the form \[\cos\theta – \tfrac{1}{2} \cos 2\theta + \tfrac{1}{3} \cos 3\theta – \dots + i(\sin\theta – \tfrac{1}{2} \sin 2\theta + \tfrac{1}{3} \sin 3\theta – \dots).\]

The real and imaginary parts are both convergent, though not absolutely convergent, unless \(\theta\) is an odd multiple of \(\pi\) ((Ex. LXXIX. 3, 4). If \(\theta\) is an odd multiple of \(\pi\) then \(z = -1\), and the series assumes the form \(-1 – \frac{1}{2} – \frac{1}{3} – \dots\), and so diverges to \(-\infty\). Thus the logarithmic series converges at all points of its circle of convergence except the point \(z = -1\).

5. The binomial series. Consider the series \[1 + mz + \frac{m(m – 1)}{2!} z^{2} + \frac{m(m – 1)(m – 2)}{3!} z^{3} + \dots.\] If $m$ is a positive integer then the series terminates. In general \[ \frac{|a_{n+1}|}{|a_{n}|} = \frac{|m – n|}{n + 1} \to 1,\] so that the radius of convergence is unity. We shall not discuss here the question of its convergence on the circle, which is a little more difficult.1


194. Uniqueness of a power series.

If \(\sum a_{n} z^{n}\) is a power series which is convergent for some values of \(z\) at any rate besides \(z = 0\), and \(f(z)\) is its sum, then it is easy to see that \(f(z)\) can be expressed in the form \[a_{0} + a_{1}z + a_{2}z^{2} + \dots + (a_{n} + \epsilon_{z})z^{n},\] where \(\epsilon_{z} \to 0\) as \(|z| \to 0\). For if \(\mu\) is any number less than the radius of convergence of the series, and \(|z| < \mu\), then \(|a_{n}| \mu^{n} < K\), where \(K\) is a constant (cf. § 192), and so \[\begin{aligned} \left|f(z) – \sum_{0}^{n} a_{\nu}z^{\nu}\right| &\leq |a_{n+1}| |z^{n+1}| + |a_{n+2}| |z^{n+2}| + \dots\\ &< K \left(\frac{|z|}{\mu}\right)^{n+1} \left(1 + \frac{|z|}{\mu} + \frac{|z|^{2}}{\mu^{2}} + \dots\right) = \frac{K |z|^{n+1}}{\mu^{n} (\mu – |z|)},\end{aligned}\] where \(K\) is a number independent of \(z\). It follows from Ex. LV. 15 that if \(\sum a_{n}z^{n} = \sum b_{n}z^{n}\) for all values of \(z\) whose modulus is less than some number \(\mu\), then \(a_{n} = b_{n}\) for all values of \(n\). This result is capable of considerable generalisations into which we cannot enter now. It shows that the same function \(f(z)\) cannot be represented by two different power series.

  1. See Bromwich, Infinite Series, pp. 225 et seq.; Hobson, Plane Trigonometry (3rd edition), pp. 268 et seq.↩︎

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