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Proof.

Homework 4, 5, 6 (a)limn an= >0. Then forn n = 2+12 we have 2n 3 4+1 3>1 >0,so 0<12n 3< , and thus|an 0|=12n 3< . (g)limn ( n+ 1 n) = >0. Then forn n =14 2we have 2 n 1 >0, and so|an 0|= n+ 1 n=( n+1 n)( n+1+ n) n+1+ n=1 n+1+ n<12 n . (k)The sequencean={1ifnis odd1/nifnis not. Then the sequence converges to some limitA R. Bydefinition of convergence (with = 1/4) there existsn such that|an A|<1/4 forn n . Choose an integerk n /2. Then 2k n and 2k+1 n ,so|a2k A|<1/4 and|a2k+1 A|<1/4. So|1/(2k) A|<1/4 and|1 A|<1/4. The second inequality impliesA>3/4, and the first oneA<1/(2k) + 1/4 1/2 + 1/4 = 3/4.}

2.1.2(k) The sequence a n = (1 if n is odd 1/n if n is even diverges. Proof. Assume not. Then the sequence converges to some limit A ∈ R. By definition of convergence (with = 1/4) ... We know that monotone bounded sequences converge, so there exists some limit A ∈ R. We can pass to the limit in the recursive equation to get

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  Sequence, Convergence, Monotone

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