# Get A course in real analysis PDF By Hugo D. Junghenn

ISBN-10: 148221928X

ISBN-13: 9781482219289

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Indeed, since an = rn → +∞, an−1 there exists N ∈ N such that an > 2an−1 , for all n > N . Iterating, we see that an > 2k an−k ≥ kan−k , so taking k = n − N we have an > (n − N )aN for all n > N . Since limn (n − N )aN = +∞ (Archimedean principle), the assertion follows. 12). 11 Theorem. Let {an } and {bn } be sequences in R. The following limit properties hold in R in the sense that if the expression on the right side of the equation exists in R, then the limit on the left side exists and equality holds.

4 Definition. The absolute value of a real number x is defined by |x| = x −x if x ≥ 0, if x < 0. ♦ For example, |0| = 0 and |2| = | − 2| = 2. 5 Proposition. Absolute value has the following properties: (a) |x| ≥ 0. (b) |x| = 0 iff x = 0. (c) | − x| = |x|. (d) − |x| ≤ x ≤ |x|. (e) |xy| = |x| |y|. (f) (g) |x + y| ≤ |x| + |y|. (h) |x| x = (y = 0). y |y| |x| − |y| ≤ |x − y|. (triangle inequalities) Proof. Properties (a)–(e) are easily established by considering cases. For example, in (e), if x ≥ 0 and y ≤ 0, then xy ≤ 0, hence |xy| = −(xy) = x(−y) = |x| |y|.

2. 2. 5. 3 This = n n + . k−1 k 8 A Course in Real Analysis 7. Without using mathematical induction, prove that for any n ∈ N, n (a) k=0 n (b) k=0 2 1 = (k + 1)(n − k + 1) n+2 n k=0 1 1 = (2k + 1)(2n − 2k + 1) n+1 1 . k+1 n k=0 1 . 3 n k=1 f (k) = n3 for all n. Order Structure of R The order relation on R is derived from the following order axiom. There exists a nonempty subset P of R, closed under addition and multiplication, such that for each x ∈ R exactly one of the following holds: x ∈ P, −x ∈ P, or x = 0.