By Daniel W. Stroock

ISBN-10: 3319244671

ISBN-13: 9783319244679

ISBN-10: 3319244698

ISBN-13: 9783319244693

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**Example text**

An } ⊆ (0, ∞) and {θ1 , . . , θn } ⊆ [0, 1] with nm=1 θm = 1, then a1θ1 . . 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.

M=0 n x n n n = ≤ m=0 |x|m m! m−1 1− n , =0 m−1 1− 1− n . =0 For any N ≥ 1 and n ≥ N , n m=0 |x|m m! m−1 N 1− 1− n ≤ =0 m=0 |x|m m! ∞ m−1 1− 1− + n m=N +1 =0 |x|m , m! and therefore, for all N ≥ 1, ∞ m=0 xm − e x = lim n→∞ m! n m=0 xm − 1− m! x n n ∞ ≤ m=N +1 |x|m . m! |x| Hence, since, by the ratio test, ∞ m=0 m! 4) after letting N → ∞. 4) can be used to estimate e. Indeed, observe that for ∞ k−m−1 = −m = any m ≥ k ≥ 2, m! k m+1−k . Hence, since ∞ m=k k m=1 k 1 k−1 , we see that n k−1 m=0 1 ≤e≤ m!

Thus if {z n : n ≥ 1} satisfies Cauchy’s criterion, so do both {xn : n ≥ 1} and {yn : n ≥ 1}. Hence, there exist x, y ∈ R such that xn → x and yn → y. Now let > 0 be given and choose n so that |xn − x| ∨ |yn − y| < √ for n ≥ n . Then 2 2 2 |z n − z|2 < 2 + 2 = 2 and therefore |z n − z| < for n ≥ n . 3 to show that every bounded sequence {z n : n ≥ 1} in C has a convergent subsequence. 46 2 Elements of Complex Analysis All the results in Sect. 2 about series and Sect. 9 about products extend more or less immediately to the complex numbers.

### A Concise Introduction to Analysis by Daniel W. Stroock

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