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Calculus 141 Section 9.6 Lecture Notes

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Calculus 141 Section 9.6 Lecture Notes Calculus 141, section 9.6 Ratio Test and Root Tests notes by Tim Pilachowski ∞ c r m ● The geometric series ∑ c r n = if and only if r <1. n = m 1 − r ∞ 1 ● The p-series converges whenever p > 1 and diverges whenever 0 < p ≤ 1. ∑ p n =1 n ∞ ∞ ● The Integral Test states a series a converges if and only if f ()x dx converges. ∑ n ∫1 n =1 ∞ ● In the Direct Comparison Test, ∑ an converges if its terms are less than those of a known convergent n =1 series, and diverges if its terms are greater than those of a known convergent series. ∞ an ● The Limit Comparison Test states: If lim exists and is a positive number, then positive series ∑ an n → ∞ bn n =1 ∞ and ∑ bn either both converge or both diverge. n =1 A downside to the Comparison Tests is that they require a suitable series to use for the comparison. In contrast, the Ratio Test and the Root Test require only the series itself. ∞ ∞ an+1 Ratio Test (Theorem 9.15) Given a positive series an for which lim = r : a. If 0 ≤ r < 1, then an ∑ n→∞ ∑ n =1 an n =1 ∞ an+1 converges. b. If r > 1, then an diverges. c. If r = 1 or lim does not exist, no conclusion. ∑ n→∞ n =1 an The proof of part a. relies upon the definition of limits and the “creation” of a geometric series which converges to which we compare our original series. Briefly, by the nature of inequalities there exists a value s for which an+1 0 ≤ r < s < 1. From the definition of limits, there exists a value N such that for n ≥ N, ≤ s ⇒ an+1 ≤ ans . an 2 n Then 0 < aN +2 ≤ aN +1s ≤ ()aN s s = aN s ⇒ 0 < aN +n ≤ aN s . The latter value is the basis for a geometric series which converges. The application of the Comparison Test finishes the proof of part a. The proof of part b. is analogous, with r > s > 1. 1 1 Hint: The Ratio Test works best for series such as ∑ , ∑ r n , and ∑ for which n is a factorial or n! 2n + c an exponent. ∞ 100n Example A: Does ∑ converge? Answer: yes n = 0 n! ∞ nn Example B: Does ∑ converge? Answer: no n =1 n! ∞ n! Example B upside down: Does the “reciprocal” series converge? Answer: yes ∑ n n =1 n ⎧ n! ⎫ ∞ ∞ n! In Lecture 9.4 it was noted that since converges to 0, although we could say that might ⎨ n ⎬ ∑ n ⎩n ⎭n = 1 n =1 n converge, we could not be certain that it does converge. The nature of the Ratio Test is such that if it shows a series converges, then the series involving the reciprocals of the terms must diverge, and vice-versa. ∞ 1 ∞ 1 Example C: Use the Ratio Test to test convergence of and . Answer: inconclusive ∑ ∑ 2 n =1 n n =1 n ∞ n Root Test (Theorem 9.16) Given a positive series ∑ an for which lim an = r : a. If 0 ≤ r < 1, then n =1 n → ∞ ∞ ∞ n ∑ an converges. b. If r > 1, then ∑ an diverges. c. If r = 1 or lim an = r DNE, no conclusion. n =1 n =1 n → ∞ The proof is a little simpler than that for the Ratio Test: For values s and N as described above, n n an ≤ s ⇒ an ≤ s , the basis for a geometric series which converges. The Comparison Test finishes the proof of part a. The proof of part b. is analogous, with r > s > 1. The Root Test is especially useful in series that involve a kth power and which have no complications such as factorials. ∞ 1 Example D: Does converge? Answer: yes ∑ n n = 2 ()ln n n2 ∞ ⎛ 1 ⎞ Example E: Does ∑ ⎜1+ ⎟ converge? Answer: no n =1⎝ n ⎠ .
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