(a) Using Cauchy's general principle of convergence, examine the convergence of the sequence < fₙ >, where fₙ = 1 + 1/1! + 1/2! +…

(a) Let fₙ = 1 + 1/1! + 1/2! + ... + 1/n! = Σ from k=0 to n of 1/k!. By Cauchy’s general principle of convergence, we must show that for every ε > 0 there exists N such that |f_m − f_n| < ε whenever m > n ≥ N.

Assume m > n. Then

|f_m − f_n| = Σ from k=n+1 to m of 1/k!. For k ≥ 1, k! = 1·2·3...k ≥ 2^(k−1), so 1/k! ≤ 1/2^(k−1). Hence

|f_m − f_n| ≤ Σ from k=n+1 to ∞ of 1/2^(k−1) = 1/2^(n−1).

Given ε > 0, choose N such that 1/2^(N−1) < ε. Then for all m > n ≥ N,

|f_m − f_n| ≤ 1/2^(n−1) ≤ 1/2^(N−1) < ε. Thus the sequence is Cauchy and therefore convergent. Its limit is the series sum e = Σ from k=0 to ∞ of 1/k!.

(b) Let G be an abelian group and let φ : G → H be a homomorphism. Its homomorphic image is φ(G). Take any u, v ∈ φ(G). Then there exist x, y ∈ G such that u = φ(x), v = φ(y). Since G is abelian, xy = yx. Therefore uv = φ(x)φ(y) = φ(xy) = φ(yx) = φ(y)φ(x) = vu. So every pair of elements of φ(G) commutes. Hence every homomorphic image of an abelian group is abelian.

The converse is not necessarily true. Take G = S₃, the symmetric group on three symbols, which is non-abelian, for example (12)(23) ≠ (23)(12). Define the sign homomorphism sgn : S₃ → {1, −1}. Its image is {1, −1}, which is cyclic of order 2 and hence abelian. Thus a non-abelian group can have an abelian homomorphic image. Therefore the converse fails.

(c) Put z = e^(iθ) on the circle C. Then cos θ = (z + z⁻¹)/2, sin θ = (z − z⁻¹)/(2i), cos 2θ = (z² + z⁻²)/2.

The given boundary value is W(θ) = [(a² − 1) cos θ + i(a² + 1) sin θ] / [a⁴ − 2a² cos 2θ + 1].

Simplify the numerator: (a² − 1) cos θ + i(a² + 1) sin θ = (a² − 1)(z + z⁻¹)/2 + (a² + 1)(z − z⁻¹)/2 = a² z − z⁻¹.

Now simplify the denominator: a⁴ − 2a² cos 2θ + 1 = a⁴ − a²(z² + z⁻²) + 1 = (a² − z²)(a² − z⁻²).

Therefore, on C, W(θ) = (a² z − z⁻¹) / [(a² − z²)(a² − z⁻²)] = z / (a² − z²).

Thus the required analytic function is f(z) = z/(a² − z²).

Since a² > 1, the zeros of the denominator are z = ±a, and |a| > 1, so they lie outside the unit circle C. Hence f(z) is analytic inside and on C. By the uniqueness theorem for analytic functions, this is the required function.

Check independently: for z = e^(iθ), f(e^(iθ)) = 1/(a² e^(−iθ) − e^(iθ)) = [(a² − 1) cos θ + i(a² + 1) sin θ] / [(a² − 1)² cos²θ + (a² + 1)² sin²θ], and the denominator equals a⁴ − 2a² cos 2θ + 1. Hence it matches the given boundary value exactly.