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.