41 prove the following: (a) lim n→∞ (3an −7) =2. (b) lim n→∞ an +1 an = 4 3 . (Hint: prove first that there is Nsuch that an >1 for n≥N.) 2.1.14 Let an ≥0 for all n∈N. Prove that if limn→∞an =ℓ, then limn→∞√an = √ℓ. 2.1.15 Prove that the sequence {an}withan =sin(nπ/2) is divergent. 2.1.16 ▶Consider a sequence {an}. (a) Prove that limn→∞an =ℓ if and only if limk→∞a2k =ℓ and limk→∞a2k+1 =ℓ. (b) Prove that limn→∞an =ℓ if and only if limk→∞a3k =ℓ, limk→∞a3k+1 =ℓ, and limk→∞a3k+2 =ℓ. 2.1.17 Given a sequence {an}, define a new sequence {bn}by bn = a1 +a2 +. . .+an n . (a) Prove that if limn→∞an =ℓ, then limn→∞bn =ℓ. (b) Find a counterexample to show that the converse does not hold in general. 2.2 Limit Theorems We now prove several theorems that facilitate the computation of limits of some sequences in terms of those of other simpler sequences. Theorem 2.2.1 Let {an}and{bn}be sequences of real numbers and let k be a real number. Suppose {an} converges to a and {bn} converges to b. Then the sequences {an +bn}, {kan}, and {anbn} converge and (i) limn→∞(an +bn) =a+b, (ii) limn→∞(kan) =ka, (iii) limn→∞(anbn) =ab, (iv) If in addition b̸=0 and bn̸ =0 for n∈N, then {an/bn}converges and limn→∞an/bn =a/b. Proof: (i) Fix anyε >0. Since {an}converges to a, there exists N1 ∈Nsuch that |an −a| < ε 2 for all n≥N1. Similarly, there exists N2 ∈Nsuch that |bn −b| < ε 2 for all n≥N2. Let N=max{N1,N2}. For any n≥N, one has |(an +bn)−(a+b)| =|(an −a)+(bn −b)| ≤ |an −a| +|bn −b| < ε 2 + ε 2 =ε.
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