Bourbaki

Exercise 2, § 3

General Topology, VIII

Let (zn)(z_n) be an infinite sequence of complex numbers zn=xn+iynz_n = x_n + i y_n such that xn≥0x_n \geq 0 for all nn. Show that if the series whose general terms are znz_n and zn2z_n^2 are convergent, then the series whose general term is zn2z_n^2 is absolutely convergent. Give an example in which this result is false when the condition xn≥0x_n \geq 0 (which can also be written −π/2≤Am⁡(zn)≤π/2-\pi/2 \leq \operatorname{Am}(z_n) \leq \pi/2) is replaced by

−(1−ε)π2≤Am⁡(zn)≤(1+ε)π2 -(1 - \varepsilon) \frac{\pi}{2} \leq \operatorname{Am}(z_n) \leq (1 + \varepsilon) \frac{\pi}{2}
no matter how small the real number ε>0\varepsilon > 0 may be. (Here Am⁡(z)\operatorname{Am}(z) denotes that radian measure of the amplitude of znz_n which belongs to the interval ]−π,+π]]-\pi, +\pi].)

Solution

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Transcribed from the printed volume. Not translated.