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MA 16200 Study Guide - Exam # 3 Fall 2011 MA 16200 Study Guide - Exam # 3 (1) Sequences; limits of sequences; Limit Laws for Sequences; Squeeze Theorem; monotone sequences; bounded sequences; Monotone Sequence Theorem. X1 Xn X1 (2) Infinite series an; sequence of partial sums sn = ak; the series an converges to s if sn ! s. n=1 k=1 n=1 (3) Special Series: X1 a n−1 2 3 ··· j j (a) Geometric Series: ar = a(1 + r + r + r + ) = − , if r < 1 (converges). n=1 1 r The Geometric Series diverges if jrj ≥ 1. X1 1 ≤ (b) p-Series: p converges when p > 1; diverges when p 1. n=1 n X1 (4) List of Convergence Tests for an. n=1 (0) Divergence Test (1) Integral Test (2) Comparison Test (3) Limit Comparison Test (4) Alternating Series Test (5) Ratio Test (6) Root Test 1 (Useful inequality: ln x < xm, for any m > > 0:37.) e (5) Strategy for Convergence/Divergence of Infinite Series. X Usually consider the form of the series an: (i) If lim an =6 0 or DNE, the use Divergence Test. n!1 X 1 X (ii) If series is a p-Series , use p-Series conclusions; if series is a Geometric series a rn−1, np use Geometric Series conclusions. P (iii) If an looks like a p-Series or Geometric series, use Comparison Test or Limit Comparison Test. (iv) If an involves factorials, use Ratio Test. th (v) If an involves n powers, use Root Test. P (vi) If an is an alternating series, use Alternating Series Test (or use Ratio/Root Test to show absolute or conditional convergence). (vii) The last resort is usually to use the Integral Test. 1 Convergence Tests X1 0 DIVERGENCE TEST: If lim an =6 0 or DNE =) an diverges n!1 n=1 1 INTEGRAL TEST: Let f be continuous, decreasing, and positive on [1; 1) and f(n) = an. Z 1 X1 (i) If f(x) dx converges =) an converges. 1 n=1 Z 1 X1 (ii If f(x) dx diverges =) an diverges. 1 n=1 X1 Z 1 (i.e., the series an converges if and only if the improper integral f(x) dx converges.) n=1 1 X X 2 COMPARISON TEST: Suppose an and bn are series with positive terms (an; bn > 0). X X (i) If an ≤ bn for all n and bn converges =) an converges. X X (ii) If an ≥ bn for all n and bn diverges =) an diverges. P P 3 LIMIT COMPARISON TEST: Suppose an and bn are series with positive terms an (an; bn > 0). If lim = c, where 0 < c < 1, then either both series converge or both n!1 bn diverge. X1 n−1 4 ALTERNATING SERIES TEST: Given an alternating series (−1) bn. 8 n=1 > bn > 0 <> X1 f g n−1 If bn is a decreasing sequence ; then the alternating series (−1) bn converges. > :> n=1 lim bn = 0 n!1 an+1 5 RATIO TEST: Let lim = L. n!1 an X1 (i) If L < 1 =) an converges. n=1 X1 (ii) If L > 1 =) an diverges. n=1 (If L = 1, then test is inconclusive.) q n 6 ROOT TEST: Let lim janj = L. n!1 X1 (i) If L < 1 =) an converges. n=1 X1 (ii) If L > 1 =) an diverges. n=1 (If L = 1, then test is inconclusive.) X1 X1 n−1 n−1 (6) Alternating series (−1) bn; estimating alternating series: if s = (−1) bn, n=1 P n=1 P j − j ≤ j j then s sn bn+1 ConditionalP convergence (i.e., an converges, but an diverges); Absolute convergence (i.e. janj converges). NOTE: absolute convergence =) convergence. X1 n (7) Power series about a: cn(x − a) ; Radius of Convergence and Interval of Convergence; n=0 usually use Ratio Test to determine ROC and IOC (for IOC don't forget to check the endpoints!) (8) Operations on power series; differentiation, integration of power series. X1 f (n)(a) (9) Taylor Series about a: f(x) = (x − a)n. n=0 n ! X1 f (n)(0) (10) Maclaurin Series: f(x) = xn. n=0 n ! Xn (k) th f (a) k (11) n -degree Taylor polynomial : Tn = (x − a) ≈ f(x), near x = a. k ! !k=0 ! 1 X k k n ! Binomial Series : (1 + x)k = xk, where = . (12) n n − n=0 k !(n k)! (13) Useful Maclaurin Series 1 X1 n 2 3 ··· − (a) − = x = 1 + x + x + x + ; valid for 1 < x < 1 1 x n=0 X1 xn x2 x3 (b) ex = = 1 + x + + + ··· ; valid for − 1 < x < 1 n=0 n ! 2 ! 3 ! X1 x2n+1 x3 x5 x7 (c) sin x = (−1)n = x − + − + ··· ; valid for − 1 < x < 1 n=0 (2n + 1) ! 3 ! 5 ! 7 ! X1 x2n x2 x4 x6 (d) cos x = (−1)n = 1 − + − + ··· ; valid for − 1 < x < 1 n=0 (2n)! 2 ! 4 ! 6 ! X1 x2n+1 x3 x5 x7 (e) tan−1 x = = x − + − + ··· ; valid for − 1 < x < 1 n=0 (2n + 1) 3 5 7 ! 1 X k k(k − 1) k(k − 1)(k − 2) (f) (1 + x)k = xk = 1 + kx + x2 + x3 + ··· ; for − 1 < x < 1 n n=0 2 ! 3 !.
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