Calculus Deconstructed: A Second Course in First-Year by Zbigniew H. Nitecki

By Zbigniew H. Nitecki

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Extra info for Calculus Deconstructed: A Second Course in First-Year Calculus (Draft version May 28, 2008)

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Xk − yk → x − y 12 see Appendix A 48 CHAPTER 2. SEQUENCES AND THEIR LIMITS 3. xk yk → xy 4. If y = 0, xk yk → x y 5. If x > 0, then for every positive integer n, √ n xk → √ n x. Proof. Proof of (1): Given ε > 0, we need to find N so that k ≥ N guarantees |(xk + yk ) − (x + y)| < ε. 1(1) give us |(xk + yk ) − (x + y)| = |(xk − x) + (yk − y)| ≤ |xk − x| + |yk − y| . One way to make sure that our desired estimate holds is by making each of the terms in this last expression less than ε/2. But we can do this: since xk → x, we can find N1 so that k ≥ N1 guarantees |xk − x| < ε 2 and since yk → y, we can also find N2 so that k ≥ N2 guarantees |yk − y| < ε .

One way to make sure that our desired estimate holds is by making each of the terms in this last expression less than ε/2. But we can do this: since xk → x, we can find N1 so that k ≥ N1 guarantees |xk − x| < ε 2 and since yk → y, we can also find N2 so that k ≥ N2 guarantees |yk − y| < ε . 2 Now, let N ≥ max{N1 , N2 }. Then k ≥ N guarantees |(xk + yk ) − (x + y)| ≤ |xk − x| + |yk − y| < ε ε + =ε 2 2 as required. ♦ Proof of (2): This is just like the preceding proof; we leave it as an exercise (Exercise 25).

D1 d2 · · · = . 4 we saw that the partial sums of this series (which are the finite decimal expressions using these digits) are non-decreasing, so to prove convergence we needed only to establish an upper bound, which in this case was easy. 5. , the partial sums k ai , Sk = k = 1, 2, ... i=0 form a non-decreasing sequence; hence such a sum either converges or diverges to +∞. Proof. Sk+1 − Sk = ak+1 ≥ 0, so Sk ↑. 6 (Comparison Test for Positive Series). If two sequences ∞ ∞ {ai }∞ i=0 and {bi }i=0 satisfy 0 ≤ ai ≤ bi for all i and the series i=0 bi converges, then the series ∞ a also converges.

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