By K. Donner
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Extra info for Extension of Positive Operators and Korovkin Theorems
Show that k=2 k converges conditionally. k2 − 1 3. 902. Use this result to ﬁnd upper and lower bounds for S. In Exercises 9–12, show that the series converges. 005. ∞ (−1)k 9. k4 k=1 4. The alternating harmonic series can be shown to converge to ln 2. Use this fact to ﬁnd the limit of the series exactly. 10. 2. Does the partial sum S15 overestimate the limit of the series? Justify your answer. ∞ k=1 5. Consider the series in Example 8. (a) Compute S50 . ∞ dx . (b) Explain why |R50 | ≤ 50 x 3 (c) Use parts (a) and (b) to ﬁnd upper and lower bounds for the limit of the series.
K→∞ (b) Give an example of a divergent series for which lim Sn k=1 k=1 ak ≤ 100 for all n ≥ 1. Explain why lim ak = 0 must be n 63. Let Hn = k=1 k→∞ k=1 H2n = 1 + ∞ b j are 58. Suppose that the partial sums of the series 2n + 3 Sn = . b j = ln n+1 j=1 n j=1 (a) Evaluate lim Sn . n→∞ (b) Does the series converge? Justify your answer. (c) Show that b j < 0 for all j ≥ 1. ∞ ak satisfy the 59. Suppose that the partial sums of the series k=1 6 ln n inequality < Sn < 3 + ne−n for all n ≥ 100. ln(n 2 + 1) (a) Does the series converge?
The disadvantage, as the harmonic series shows, is that the partial sums may tend to inﬁnity. Mixing positive and negative terms may cost something in simplicity, but it is an advantage for convergence. As the alternating harmonic series shows, positive and negative terms can offset each other, thus helping the cause of convergence. A bs o lut e c o n v e r g e n c e i m p l i e s o r d i n a r y c o n v e r g e n c e We saw in Chapter 10 that if ∞ ∞ ∞ ∞ 1 | f (x)| d x converges, then so must 1 f (x) d x, and 1 f (x) d x ≤ 1 | f (x)| d x.
Extension of Positive Operators and Korovkin Theorems by K. Donner