We saw in § 170 that if un and vn are two convergent series of positive terms, then un×vn=wn, where wn=u0vn+u1vn1++unv0. We can now extend this result to all cases in which un and vn 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 anzn, bnzn, then the product of the two series is cnzn, where cn=a0bn+a1bn1++anb0.

2. If the radius of convergence of anzn 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)/(1z)=snzn, where sn=a0+a1++an.

3. Prove, by squaring the series for 1/(1z), that 1/(1z)2=1+2z+3z2+ if |z|<1.

4. Prove similarly that 1/(1z)3=1+3z+6z2+, the general term being 12(n+1)(n+2)zn.

5. The Binomial Theorem for a negative integral exponent. If |z|<1, and m is a positive integer, then 1(1z)m=1+mz+m(m+1)1c2˙z2++m(m+1)(m+n1)12nzn+.

[Assume the truth of the theorem for all indices up to m. Then, by Ex. 2, 1/(1z)m+1=snzn, where sn=1+m+m(m+1)12++m(m+1)(m+n1)12n=(m+1)(m+2)(m+n)12n, as is easily proved by induction.]

6. Prove by multiplication of series that if f(m,z)=1+(m1)z+(m2)z2+, 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 zn in the product series is (mn)+(m1)(mn1)+(m2)(mn2)++(mn1)(m1)+(mn).

This is a polynomial in m and m: but when m and m are positive integers this polynomial must reduce to (m+mk) 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+z22!+ 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 un=znn!, vn=znn!, then wn=(z+z)nn!.]

8. If C(z)=1z22!+z44!,S(z)=zz33!+z55!, then C(z+z)=C(z)C(z)S(z)S(z),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 un and vn are not absolutely convergent may be seen by considering the case in which un=vn=(1)nn+1. Then wn=(1)nr=0n1(r+1)(n+1r). But (r+1)(n+1r)12(n+2), and so |wn|>(2n+2)/(n+2), which tends to 2; so that wn is certainly not convergent.


191–194. Power series Main Page MISCELLANEOUS EXAMPLES ON CHAPTER VIII