By Steven G. Krantz

Do not get me incorrect - Krantz is sweet yet this is often basically child Rudin - with out the proofs - that is type of like a bar with out beer.

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A E U n X and b E V n X. Set a = sup (U n (a, bJ). b] is disjoint from V. ButCV is closed; hence a ¢ V. It follows that a < b.

3. The naive method for defining the summands of the product series is to let Cj = aj • bj. However. a glance at the producl of two partial sums of the given series shows that such a definition would be ignoring lhe distributivily of multiplication over addition. 3 The Cauchy Product Cauchy's idea was that the terms for the product series should be rn Cm :; La j . bm - j • j=O This particular form for the summands can be easily motivated using power series considerations (which we shall provide laler on).

Thus 0< Oli :: J2/{j - 1) as long as j ? 2. 3 that Oli ~ 0 or j1/i -+ 1. 17 LetOl be a positive real number. Then the sequence Oli/i converges to 1. To see this. first note that the case Ol = 1 is trivial. and the case Ol > I implies the case Ol < 1 (by taking reciprocals). So we concentrate on Ol > 1. But then we have 1 < Olili < jlli when j > Ol. Since j1/i tends to 1. 3 applies and the argument 0 is complete. 18 Let A > 1 and let Ol be real. Then the sequence {~~CI converges to O. To see this.