The fundamental limiting processes for sequences of real numbers can readily be generalized to complex numbers without even the necessity of reformulating definitions and proofs. ¹ 

1.2.1 Limit of a Sequence

An infinite sequence of complex numbers \[z_{1}, z_{2}, \ldots, z_{n}, \ldots\] is said to possess the limit \(z\) , \[\lim _{n \rightarrow \infty} z_{n}=z \quad \text { or } \quad x_{n}+i y_{n}=z_{n} \rightarrow z=x+i y\] if for any preassigned positive \(\varepsilon\) , no matter how small, there is an integer \(N=N(\varepsilon)\) such that \(\left|z_{n}-z\right|<\varepsilon\) for all \(n\) greater than \(N\) . Loosely speaking, this statement means that we can come as close to \(z\) as we please provided we go out sufficiently far in the sequence. We do not mean to exclude by this mode of expression the possibility that some or all of the \(z_{n}\) are equal. In the latter case \(z_n\) clearly coincides with \(z\) . The above definition is, of course, completely equivalent to the stipulation that the real part \(x_{n}\) and the imaginary part \(y_{n}\) of \(z_{n}\) converge separately to \(x\) and \(y\) respectively.

On the other hand, if \(\left|z_{n}\right|\) increases indefinitely in that the modulus \(\left|z_{n}\right|\) eventually remains larger than any preassigned positive value \(1 / \varepsilon\) we say that \(z_{n}\) tends to infinity and write \[z_{n} \rightarrow \infty\] This condition is tantamount to the requirement that \(1/z_{n} \rightarrow0\) . A sequence which possesses a limit is said to be convergent . The fact that a sequence is convergent is plainly an intrinsic property of the sequence which does not depend upon knowing the limiting value. It should therefore be of value to give a definition of convergence which frees us from the necessity of furnishing a specific limit. Such a definition has been provided by Cauchy:

To prove the equivalence of the two definitions of convergence we must employ the fundamental theorem of Bolzano-Weierstrass. The Bolzano-Weierstrass theorem ² states: Every bounded infinite set of points of the complex plane possesses at least one accumulation point.

Consider a sequence \[\left\{z_{n}\right\}=z_{1}, z_{2}, \ldots, z_{n}, \ldots\] of complex numbers which converges to a limit \(z\) . Geometrically this means that the points \(z_{n}\) of the sequence, from a certain \(N\) on, lie inside a circle of given radius \(\varepsilon\) about the point \(z\) . One recognizes immediately that if \(\lim z_{n}=z\) then the set of points \(\left\{z_{n}\right\}\) is bounded and \(z\) is its only accumulation point. Conversely, if a bounded infinite sequence of points \(\left\{z_{n}\right\}\) possesses only one accumulation point \(z\) then \(z\) is the limit of the sequence \(\left\{z_{n}\right\}\) . (If a sequence contains terms \(z_{n}\) equal to \(z\) these statements are interpreted by counting the point corresponding to \(z\) as often as it occurs.)

We say that a sequence “tends to the point at infinity” if all its points \(z_{n}\) , from a certain \(N\) on lie outside a circle of arbitrarily large preassigned radius \(R\) . The reason becomes apparent if we consider the points \(z_{n}\) on the \(z\) -sphere (by means of stereographic projection), for there the corresponding sequence tends to the north pole. Nevertheless, when we say without further qualification that a sequence converges we mean that it possesses a finite limit.

The equivalence of the two definitions of convergence is proved as follows:

1. A sequence which converges to a limit is convergent in the sense of Cauchy. For suppose \(z_{n} \rightarrow z\) . Then, given an \(\varepsilon>0\) , there exists an \(N\) such that \[\left|z_{n}-z\right|<\frac{\varepsilon}{2},\quad \left|z_{m}-z\right|<\frac{\varepsilon}{2}\quad \text {for } n, m > N.\] Then, according to the triangle inequality, we have \begin{align} \left|z_{n}-z_{m}\right|&=\left|\left(z_{n}-z\right)+\left(z-z_{m}\right)\right|\\ &\leq\left|z_{n}-z\right|+\left|z_{m}-z\right|\\ &<\frac{\varepsilon}{2}+\frac{\varepsilon}{2} =\varepsilon\quad \text{for } n, m > N. \end{align} 
2. Conversely, any sequence convergent in the sense of Cauchy tends to a definite limit. Let \(\left\{z_{n}\right\}\) be a Cauchy sequence. From (2.11) it follows for a given arbitrarily small \(\varepsilon\) that there exists an \(N\) such that \[\left|z_{N+1}-z_{m}\right|<\varepsilon\quad \text {for all } m>N,\] i.e., all the points after \(z_{N+1}\) lie within a circle of radius \(\varepsilon\) about \(z_{N+1}\) , while outside this circle there are only a finite number of points of the sequence – at most \(z_{1}, \ldots, z_{N}\) . Thus the sequence is a bounded point set and, by the Bolzano-Weierstrass theorem, possesses at least one accumulation point. This will prove that the sequence tends to a limit. Suppose then that there are two accumulation points, say \(z\) and \(z^{*}\) . This means that there are infinitely many numbers \(z_{n}\) arbitrarily near \(z\) as well as infinitely many numbers \(z_{m}\) arbitrarily near \(z^{*}\) , i.e. \[\left|z_{m}-z\right|<\frac{\varepsilon}{3}, \quad\left|z_{n}-z^{*}\right|<\frac{\varepsilon}{3}\] for infinitely many numbers \(n\) and \(m\) , which, together with the Cauchy condition gives \[\left|z_{m}-z_{n}\right|<\frac{\varepsilon}{3}\quad \text {for } m, n> N\left(\frac{\varepsilon}{3}\right)\] gives \[|z-z^*| \leq\left|z-z_{m}\right|+\left|z_{m}-z_{n}\right|+\left|z_{n}-z^{*}\right|<\varepsilon.\] Since, on the other hand, \(\varepsilon\) can be chosen arbitrarily small, it follows that \(z\) and \(z^{*}\) are identical.

The Cauchy condition (2.11) is a necessary and sufficient condition for a sequence to converge to a limit. 

1.2.2 Infinite Series

A series of complex numbers \[\sum_{\nu=1}^{\infty} a_{\nu}=a_{1}+a_{2}+\cdots+a_{n}+\cdots\] is said to be convergent or to have the sum \(S\) if the sequence \(\left\{S_{n}\right\}\) of partial sums \[S_{n}=\sum_{\nu=1}^{n} a_{\nu}=a_{1}+\cdots+a_{n}\] converges to a limit \[S=\sum_{\nu=1}^{\infty} a_{\nu}=\lim S_{n}.\] If a series does not possess a sum in this sense it is said to be divergent . A necessary and sufficient condition for the convergence of a series is provided by the Cauchy criterion (2.11) , namely, for any positive \(\varepsilon\) there is an \(N(\varepsilon)\) such that \[\tag{2.21} \left|S_{n} - S_{n+p}\right|=\left|a_{n+1}+a_{n+2}+\cdots+a_{n+p}\right|<\varepsilon\] for all \(n>N(\varepsilon)\) , \(p \geq 1\) . Hence for any convergent series \(\left|a_{n}\right| \rightarrow 0\) . This condition alone will not guarantee the existence of a sum since e.g. the harmonic series \[1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}+\cdots\] does not converge.

In any specific instance it may not be easy to prove convergence by a direct application of the Cauchy condition. For this reason we introduce various sufficient conditions for convergence called convergence tests which, in a great many instances, enable us to show convergence with relative ease.

Absolute Convergence

A series \(\sum_{\nu=1}^{\infty} z_{\nu}\) is said to converge absolutely if the series \(\sum_{\nu = 1}^{\infty}|z_{\nu}|\) converges. Absolute convergence implies ordinary convergence, since for any \(\varepsilon>0\) we can find an \(N(\varepsilon)\) such that \begin{align} \varepsilon > \big||z_{n+1}|+\cdots+|z_{n+p}|\big| &= |z_{n+1}|+\cdots+|z_{n+p}|\\ &\geq|z_{n+1}+\cdots+z_{n+p}| \end{align} for all \(n>N(\varepsilon)\) , \(p \geq 1\) . The converse is not true, for the series \[1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots\] converges but not absolutely. A series is called conditionally convergent if it converges but does not converge absolutely.

Comparison Test

If a series \(\sum_{\nu=1}^{n} a_{\nu}\) converges absolutely then any series \(\sum_{\nu=1}^{n} a_{\nu}'\) with \(|a_{n}'| \leq |a_{n}|\) for all \(n>N\) must also converge absolutely. The proof follows directly from the inequality \begin{align} |a_{n+1}^{\prime}+\cdots+a_{n+p}^{\prime}| &\leq |a_{n+1}^{\prime}|+\cdots+|a_{n+p}^{\prime}|\\ &\leq |a_{n+1}|+\cdots+|a_{n+p}|. \end{align} A most useful series for comparison is the geometric series \[\sum_{\nu=0}^{\infty} r^{\nu}=1+r+r^{2}+\cdots+r^{n}+\cdots\] which can be proved to converge for \(|r|<1\) to the value \(\frac{1}{1-r}\) and to diverge for \(|r|>1\) . Employing this series we obtain a number of interesting convergence criteria: if there is a constant \(r\) such that \[\tag{2.22} \left|\frac{a_{n+1}}{a_{n}}\right| \leq r<1\] for all \(n \geq N\) then the series \(\sum_{\nu=0}^{\infty} a_{\nu}\) converges absolutely. For the proof we disregard the sum of the earlier terms \[S_{N-1} = a_{0}+ a_{1} + \cdots+ a_{N-1}\] since this can affect the sum only by adding a constant. Therefore, without losing generality, we take \(N = 0\) . It follows from (2.22) that \[\left|a_{n+1}\right| \leq r\left|a_{n}\right| \leq r^{2}\left|a_{n-1}\right| \leq \cdots \leq r^{n+1}\left|a_{0}\right|.\] By comparison with the geometric series we see that this series converges absolutely.

On the other hand, if \[\tag{2.22a} \left|\frac{a_{n+1}}{a_{n}}\right| \geq 1\] for all \(n \geq N\) the series diverges. As above we may take \(N=0\) . Then \(\left|a_{n}\right| \geq\left|a_{0}\right|\) whence \(\left|a_{n}\right|\) does not tend to zero so that the series cannot possibly converge.

By a similar comparison it is easy to show that the series \(\sum_{\nu=1}^{\infty} a_{\nu}\) converges absolutely if \[\tag{2.23} \sqrt[n]{\left|a_{n}\right|} \leq r<1\] for all \(n>N\) . Again, the series \(\sum_{\nu=1}^{\infty} a_{\nu}\) diverges \[\tag{2.23a} \sqrt[n]{\left|a_{n}\right|} \geq 1\] for all terms after a given term. In fact, for divergence it is obviously sufficient just to require (2.23a) for an infinity of values of \(n\) .

In (2.22) and (2.23) it is quite essential that \(\left|\frac{a_{n+1}}{a_{n}}\right|\) and \(\sqrt[n]{\left|a_{n}\right|}\) should be bounded away from unity. It is not enough to have, say \[\left|\frac{a_{n+1}}{a_{n}}\right|<1\] for already in the harmonic series \(\sum_{n=1}^{\infty} \frac{1}{n}\) we have \[\left|\frac{a_{n+1}}{a_{n}}\right|=\frac{n}{n+1}=1-\frac{1}{n+1}<1.\] 

The convergence tests (2.22) and (2.23) may be rewritten employing the notions of superior limit and inferior limit. The superior limit of a bounded infinite set of real numbers \(x\) , denoted by \(\overline{\lim}\,x\) or \(\limsup\,x\) is the greatest of the accumulation points of the set. The superior limit always exists. For proof consider the set of values \(y\) which are exceeded by at most a finite number of the values \(x\) . Since this set is bounded below it possesses a greatest lower bound; call it \(\lambda\) . \[\lambda=\overline{\lim}\,x\] 

1. \(\lambda\) is a point of accumulation for the set of numbers \(x\) . For consider any \(\varepsilon\) -interval \(\lambda-\varepsilon<\xi<\lambda+\varepsilon\) about \(\lambda\) . From the definition of \(\lambda\) it follows that there are infinitely many values of \(x>\lambda-\varepsilon\) and at most a finite number of values of \(x \geq \lambda+\varepsilon\) . We conclude that any interval about \(\lambda\) must contain an infinity of points of the set.
2. \(\lambda\) is the greatest of the points of accumulation of the set \(x\) . For suppose \(\lambda'>\lambda\) is another. Take \(\varepsilon=\frac{\lambda'-\lambda}{2}\) . Since \(\lambda'\) is an accumulation point there are an infinite number of values of \(x\) in the interval \(\lambda'-\varepsilon<\xi<\lambda'+\varepsilon\) . But this – implies that there are infinitely many values of \(x>\lambda+\varepsilon=\lambda'-\varepsilon\) contrary to the definition of \(\lambda\) .

If the set of values of \(x\) has no upper bound we adopt the special notation \[\overline{\lim}\,x = \infty.\] In precisely the same manner as above we may define \(\underline{\lim}\,x\) . More simply we set \[\underline{\lim}\,x = -\overline{\lim}\,(-x).\] 

From (2.22) and (2.23) it follows that the D’Alembert convergence test \[\tag{2.24} \overline{\lim}\left|\frac{a_{n+1}}{a_{n}}\right|<1\] and the Cauchy convergence test \[\tag{2.25} \overline{\lim}\left|\sqrt[n]{\left|a_{n}\right|}\right|<1\] are both sufficient conditions that the series \(\sum_{\nu=1}^{\infty} a_{\nu}\) converge absolutely. For if, say \(\lambda=\overline{\lim} \sqrt[n]{\left|a_{n}\right|}<1\) then there is an \(\varepsilon>0\) such that \(r=\lambda+\varepsilon<1\) . Since there are only a finite number of values of \(n\) for which \(\sqrt[n]{\left|a_{n}\right|}>r=\lambda+\varepsilon\) convergence follows from (2.23) . Similarly, it is possible to prove the complementary results that \[\tag{2.26} \underline{\lim}\left|\frac{a_{n+1}}{a_{n}}\right|>1\] and \[\tag{2.27} \underline{\lim} \sqrt[n]{\left|a_{n}\right|}>1\] are sufficient to show divergence. According to the remark after (2.23a) we may replace (2.27) by the weaker condition \[\tag{2.28} \overline{\lim} \sqrt[n]{\left|a_{n}\right|}>1\] 

Exercises

1.2.3 Power Series

A complex function \(f(z)\) defined over a point set \(D\) of the \(z\) -plane associates with each point of \(D\) a complex value \(f(z)\) . The classical theory of functions can be treated from a point of view due principally to Weierstrass as the study of functions that can be represented by a convergent power series. In this section we shall investigate a number of the important properties of these series.

Consider a series of functions \(\sum_{\nu = 1}^{\infty} u_{\nu}(z)\) convergent for all values of \(z\) in a point set \(D\) of the complex plane. The series is said to converge uniformly in \(D\) if for every \(\varepsilon>0\) , however small, it is possible to find an \(N(\varepsilon)\) independently of \(z\) so that for all \(n > N, p \geq 1\) \[\left|u_{n+1}(z)+u_{n+2}(z)+\cdots+u_{n+p}(z)\right|<\varepsilon\] for all \(z\) in \(D\) .

If the terms of the series \(\sum_{\nu = 1}^{\infty} u_{\nu}(z)\) satisfy the condition \(\left|u_{\nu}(z)\right| \leq c_{\nu}\) for all \(z\) in \(D\) and the \(c_{\nu}\) s are real constants such that \(\sum_{\nu = 1}^{\infty} c_{\nu}\) converges then \(\sum_{\nu = 1}^{\infty} u_{\nu}(z)\) converges uniformly, and, of course, absolutely in \(D\) .

For, given any \(\varepsilon\) , it is possible to find an \(N\) sufficiently large so that \begin{align} \varepsilon>c_{n+1}+\cdots+c_{n+p} &\geq\left|u_{n+1}(z)\right|+\cdots+\left|u_{n+p}(z)\right|\\ &\geq\left|u_{n+1}(z)+\cdots+u_{n+p}(z)\right| \end{align} for all \(n > N\) , \(p \geq 1\) .

A power series is a series of the form \[\tag{2.31} P(z)=\sum_{\nu=0}^{\infty} a_{\nu} z^{\nu}=a_{0}+a_{1} z+\cdots+a_{n} z^{n}+\cdots\] where the coefficients \(a_{0}, a_{1}, a_{2}, \ldots\) may be any complex numbers.

The principal theorem on power series states:

The series \[\tag{2.32} D(z)=\sum_{\nu=1}^{\infty} \nu a_{\nu} z^{\nu-1}\] also converges if \(|z| \leq r|\xi|\) . For we have \[\left|n a_{n} z^{n-1}\right|Comparison with the series of Exercise 19 , gives the desired result.

There are two possibilities for any power series: Either it converges for all values of \(z\) or there is at least one value \(z=\eta\) for which the series diverges. In the latter case the series must diverge for every value of \(z\) for which \(|z|>|\eta|\) . If the series were convergent for such a value of \(z\) then by the theorem above it would converge for \(z=\eta\) .

We conclude that a power series convergent for some \(z \neq 0\) and divergent for some other value has a radius of convergence \(\rho\) so that the series converges absolutely for \(|z|<\rho\) and diverges for \(|z|>\rho\) . The circle \(|z|=\rho\) is called the circle of convergence of the series.

The series \(D(z)\) clearly converges inside the circle of convergence of \(P(z)\) . We prove further that \(D(z)\) diverges for \(|z| > \rho\) . There is a \(p > 1\) , \(0 < q < \rho\) such that \(|z| > pq > \rho\) . ³ Since \(q < \rho\) , \(\sum_{\nu = 1}^\infty a_\nu q^\nu\) converges and therefore \(a_n q^n\) is bounded, \(|a_n q^n| < M\) for all \(n\) . Now \begin{align} |na_n z^{n-1}| > n\,\frac{M}{q}\,p^{n-1} &\geq \left(\frac{M}{q}\right)\,p^{n-1}\\ &> \frac{M}{q}\\ &> 0. \end{align} It follows that the \(n^\text{th}\) term does not tend to zero and therefore \(D(z)\) diverges for \(|z| > \rho\) .

The theorem is a direct consequence of the tests (2.25) and (2.28) . In particular, we note that a series converges everywhere if \(\lambda=0\) , and converges only at \(z=0\) if \(\lambda=\infty\) .

No general statement can be made about the behavior of a power series on the circle of convergence. However, if the series is absolutely convergent for some value on the circle then it is plainly convergent everywhere on the circle.

Differentiation of Power Series

It is natural to call a complex function of the form \[f(z)=a_{0}+a_{1} z+a_{2} z^{2}+\cdots+a_{n} z^{n} \quad\left(a_{n} \neq 0\right)\] with fixed complex coefficients \(a_{\nu}\) , a polynomial of \(\boldsymbol{n}^{\text{th}}\) degree in \(\boldsymbol{z}\) . A convergent power series \[P(z)=\sum_{\nu=0}^{\infty} a_{\nu} z^{\nu}\] may be regarded as a function of \(z\) , defined in its circle of convergence to be the limit of the sequence of polynomials \[P_{n}(z)=\sum_{\nu=0}^{\infty} a_{\nu} z^{\nu}.\] 

The derivative of a complex function \(f(z)\) is defined exactly as in the real case as ⁴ 

\[\lim _{\zeta \rightarrow z} \frac{f(\zeta)-f(z)}{\zeta-z}.\] First we note the algebraic identity \[\frac{\zeta^{n}-z^{n}}{\zeta-z}=\zeta^{n-1}+\zeta^{n-2} z+\cdots+z^{n-1}.\] If we now let \(\zeta\) tend to \(z\) we immediately have \[\frac{d}{d z} z^{n}=\lim _{\zeta \rightarrow z} \frac{\zeta^{n}-z^{n}}{\zeta-z}=n z^{n-1}.\] It follows that \begin{align} P_{n}'(z)=\frac{d}{d z} P_{n}(z)&=\lim _{\zeta \rightarrow z} \frac{P_{n}(\zeta)-P_{n}(z)}{\zeta-z}\\ &= \sum_{\nu=1}^{n} \nu a_{\nu} a^{\nu-1}\\ &=D_{n}(z) \end{align} where \(P_{n}'(z)\) and \(\dfrac{d}{d z} P_{n}(z)\) denote the derivative of the complex polynomial \(P_{n}(z)\) .

We now derive a theorem fundamental in the theory of power series:

Since the derivative of a power series is again a power series with the same circle of convergence we can differentiate again and repeat the process as often as we like. It follows that a power series possesses derivatives of all orders within its circle of convergence .

The series \(P(z)=\sum_{\nu=0}^{\infty} a_{\nu} z^{\nu}\) is the derivative of the power series \[\tag{2.34} I(z)=\sum_{\nu=0}^{\infty} \frac{a_{\nu}}{\nu+1} z^{\nu+1}.\] We may therefore regard the series \(I(z)\) as an indefinite integral of \(P(z)\) . Clearly \(I(z)\) possesses the same circle of convergence as its derivative \(P(z)\) . Thus, it is a simple extension of our results that a power series may be integrated within its circle of convergence as often as we please.

Exercises