By Jiri Lebl

A primary path in mathematical research. Covers the true quantity procedure, sequences and sequence, non-stop features, the spinoff, the Riemann critical, sequences of capabilities, and metric areas. initially built to coach Math 444 at college of Illinois at Urbana-Champaign and later more suitable for Math 521 at college of Wisconsin-Madison. See http://www.jirka.org/ra/

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**Additional resources for Basic Analysis: An Introduction to Real Analysis**

**Example text**

Suppose such an M exists, then for an even n ≥ M we compute 1/2 > |xn − x| = |1 − x| 1/2 > |xn+1 − x| = |−1 − x| . and But 2 = |1 − x − (−1 − x)| ≤ |1 − x| + |−1 − x| < 1/2 + 1/2 = 1, and that is a contradiction. 1. 6. A convergent sequence has a unique limit. The proof of this proposition exhibits a useful technique in analysis. Many proofs follow the same general scheme. We want to show a certain quantity is zero. We write the quantity using the triangle inequality as two quantities, and we estimate each one by arbitrarily small numbers.

As r − L > 0, n| there exists an M ∈ N such that for all n ≥ M we have |xn+1 | − L < r − L. |xn | Therefore, |xn+1 | < r. |xn | For n > M (that is for n ≥ M + 1) we write |xn | = |xM | |xn | |xn−1 | |xM+1 | ··· < |xM | rr · · · r = |xM | rn−M = (|xM | r−M )rn . |xn−1 | |xn−2 | |xM | The sequence {rn } converges to zero and hence |xM | r−M rn converges to zero. 10, the M-tail of {xn } converges to zero and therefore {xn } converges to zero. Now suppose L > 1. Pick r such that 1 < r < L. As L − r > 0, there exists an M ∈ N such that for all n ≥ M we have |xn+1 | − L < L − r.

Set nk+1 := m. The subsequence {xnk } is defined. Next we need to prove that it has the right limit. ) and that ank ≥ xnk . Therefore, for every k > 1 we have |ank − xnk | = ank − xnk ≤ a(nk−1 +1) − xnk 1 < . k Let us show that {xnk } is convergent to x. Note that the subsequence need not be monotone. Let ε > 0 be given. As {an } converges to x, then the subsequence {ank } converges to x. Thus there exists an M1 ∈ N such that for all k ≥ M1 we have ε |ank − x| < . 2 Find an M2 ∈ N such that 1 ε ≤ .