We saw in § 170 that if $$\sum u_{n}$$ and $$\sum v_{n}$$ are two convergent series of positive terms, then $$\sum u_{n} \times \sum v_{n} = \sum w_{n}$$, where $w_{n} = u_{0}v_{n} + u_{1}v_{n-1} + \dots + u_{n}v_{0}.$ We can now extend this result to all cases in which $$\sum u_{n}$$ and $$\sum v_{n}$$ are absolutely convergent; for our proof was merely a simple application of Dirichlet’s Theorem, which we have already extended to all absolutely convergent series.

Example LXXXI
1. If $$|z|$$ is less than the radius of convergence of either of the series $$\sum a_{n}z^{n}$$, $$\sum b_{n}z^{n}$$, then the product of the two series is $$\sum c_{n}z^{n}$$, where $$c_{n} = a_{0}b_{n} + a_{1}b_{n-1} + \dots + a_{n}b_{0}$$.

2. If the radius of convergence of $$\sum a_{n}z^{n}$$ is $$R$$, and $$f(z)$$ is the sum of the series when $$|z| < R$$, and $$|z|$$ is less than either $$R$$ or unity, then $$f(z)/(1 – z) = \sum s_{n}z^{n}$$, where $$s_{n} = a_{0} + a_{1} + \dots + a_{n}$$.

3. Prove, by squaring the series for $$1/(1 – z)$$, that $$1/(1 – z)^{2} = 1 + 2z + 3z^{2} + \dots$$ if $$|z| < 1$$.

4. Prove similarly that $$1/(1 – z)^{3} = 1 + 3z + 6z^{2} + \dots$$, the general term being $$\frac{1}{2}(n + 1)(n + 2)z^{n}$$.

5. The Binomial Theorem for a negative integral exponent. If $$|z| < 1$$, and $$m$$ is a positive integer, then $\frac{1}{(1 – z)^{m}} = 1 + mz + \frac{m(m + 1)}{1c\dot2} z^{2} + \dots + \frac{m(m + 1) \dots (m + n – 1)}{1\cdot2 \dots n} z^{n} + \dots.$

[Assume the truth of the theorem for all indices up to $$m$$. Then, by Ex. 2, $$1/(1 – z)^{m+1} = \sum s_{n}z^{n}$$, where \begin{aligned} s_{n} &= 1 + m + \frac{m(m + 1)}{1\cdot2} + \dots + \frac{m(m + 1) \dots (m + n – 1)}{1\cdot2 \dots n} \\ &= \frac{(m + 1)(m + 2) \dots (m + n)}{1\cdot2 \dots n},\end{aligned} as is easily proved by induction.]

6. Prove by multiplication of series that if $f(m, z) = 1 + \binom{m}{1} z + \binom{m}{2} z^{2} + \dots,$ and $$|z| < 1$$, then $$f(m, z)f(m’, z) = f(m + m’, z)$$. [This equation forms the basis of Euler’s proof of the Binomial Theorem. The coefficient of $$z^{n}$$ in the product series is $\binom{m’}{n} + \binom{m}{1} \binom{m’}{n – 1} + \binom{m}{2} \binom{m’}{n – 2} + \dots + \binom{m}{n – 1} \binom{m’}{1} + \binom{m}{n}.$

This is a polynomial in $$m$$ and $$m’$$: but when $$m$$ and $$m’$$ are positive integers this polynomial must reduce to $$\dbinom{m + m’}{k}$$ in virtue of the Binomial Theorem for a positive integral exponent, and if two such polynomials are equal for all positive integral values of $$m$$ and $$m’$$ then they must be equal identically.]

7. If $$f(z) = 1 + z + \dfrac{z^{2}}{2!} + \dots$$ then $$f(z)f(z’) = f(z + z’)$$. [For the series for $$f(z)$$ is absolutely convergent for all values of $$z$$: and it is easy to see that if $$u_{n} = \dfrac{z^{n}}{n!}$$, $$v_{n} = \dfrac{z’^{n}}{n!}$$, then $$w_{n} = \dfrac{(z + z’)^{n}}{n!}$$.]

8. If $C(z) = 1 – \frac{z^{2}}{2!} + \frac{z^{4}}{4!} – \dots,\quad S(z) = z – \frac{z^{3}}{3!} + \frac{z^{5}}{5!} – \dots,$ then $C(z + z’) = C(z)C(z’) – S(z)S(z’),\quad S(z + z’) = S(z)C(z’) + C(z)S(z’),$ and $\{C(z)\}^{2} + \{S(z)\}^{2} = 1.$

9. Failure of the Multiplication Theorem. That the theorem is not always true when $$\sum u_{n}$$ and $$\sum v_{n}$$ are not absolutely convergent may be seen by considering the case in which $u_{n} = v_{n} = \frac{(-1)^{n}}{\sqrt{n + 1}}.$ Then $w_{n} = (-1)^{n} \sum_{r=0}^{n} \frac{1}{\sqrt{(r + 1)(n + 1 – r)}}.$ But $$\sqrt{(r + 1)(n + 1 – r)} \leq \frac{1}{2}(n + 2)$$, and so $$|w_{n}| > (2n + 2)/(n + 2)$$, which tends to $$2$$; so that $$\sum w_{n}$$ is certainly not convergent.