By Daniel W. Stroock

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**Example text**

Anθn ≤ n θm a m . m=1 When θm = 1 n for all m, this is the classical arithmetic-geometric mean inequality. 18 Show that exp grows faster than any power of x in the sense that α lim x→∞ xe x = 0 for all α > 0. Use this to show that log x tends to infinity more x slowly than any power of x in the sense that lim x→∞ log x α = 0 for all α > 0. Finally, α show that lim x 0 x log x = 0 for all α > 0. 19 Show that gent. 20 Just as is the case for absolutely convergent series (cf. 5), absolutely convergent products have the property that their products do not depend on the order in which their factors of multiplied.

Bn−1 ∈ C. In particular, there can be no more than n distinct solutions to f (z) = 0. (See the fundamental theorem of algebra in Sect. 2 for more information). m Proof If f (0) = 0, then a0 = 0, and so f (z) = z n−1 m=0 am+1 z . Next suppose that f (ζ) = 0 for some ζ = 0, and consider f ζ (z) = f (z + ζ). By the binomial theorem, f ζ is again an nth order polynomial, and clearly f ζ (0) = 0. Hence f (z + ζ) = zgζ (z), where gζ is a polynomial of order (n − 1), which means that we can take g(z) = gζ (z − ζ).

1) is the extension of log to C \ {0}. For this purpose, think of log on (0, ∞) as the inverse of exp on R. 1) tells us how to extend log to C \ {0} as the inverse of exp on C. Indeed, given z = 0, set r = |z| and take θ to be the unique element of (−π, π] for which z = r eiθ . Then, if we take log z = log r + iθ, it is obvious that elog z = z.