<p> Calc 2 Lecture Notes Section 8.3 Page 1 of 5</p><p>Which test do I use for a sequence or series?</p><p>Are you evaluating a sequence or a series (infinite sum)? If a series, go to page 3. Otherwise…</p><p>For a Sequence Calculate a few terms Graph some terms</p><p>Definition 1.1: Convergence of a sequence and divergence of a sequence.</p><p>The set {an} converges to L if and only if given any number  > 0 there is an integer N for n= n0</p><p> which an - L < e for every n > N. If there is no such number L, then we say the sequence diverges.</p><p>Theorem 1.1: Combinations of limits of sequences.</p><p>If {an} and {bn} both converge, then n= n0 n= n0</p><p> lim(an+ b n) = lim a n + lim b n (i). n n  n </p><p> lim(an- b n) = lim a n - lim b n (ii). n n  n </p><p>(iii). lim(an b n) = lim a n lim b n n( n )( n  )</p><p> lim an 骣an n (iv). lim 琪 = (assuming limbn 0). n n 桫bnlim b n n</p><p>Theorem 1.2: The limit of a sequence is the limit of its function. lim f( x) = L lim f( n) = L If x and x , then n for n . lim cos( 2p n) Note that the converse is not true. Counterexample: n</p><p>Theorem 1.3: Squeeze Theorem</p><p>Suppose that {an} and {bn} both converge to the limit L. If there is an integer n1  n0 such n= n0 n= n0</p><p> that all n  n1 guarantees that an  cn  bn, then {cn} converges to L as well. n= n0</p><p>Corollary 1.1: </p><p> liman = 0 liman = 0 If n , then n as well. Calc 2 Lecture Notes Section 8.3 Page 2 of 5</p><p>Definition 1.3: Increasing and Decreasing sequences. a a＃ a a ＃ a ＃ a  The sequence { n}n=1 is increasing if 1 2 3n n+ 1 a a吵 a a 吵 a 吵 a  The sequence { n}n=1 is decreasing if 1 2 3n n+ 1 If a sequence is either increasing or decreasing it is called monotonic.</p><p>Key trick: To determine whether a series is monotonic, look at the ratio of successive terms.</p><p>Definition 1.4: Bounded sequences. a The sequence { n}n=1 is bounded if there is a number M > 0 (called a bound) for which |an| < M for all n.</p><p>Theorem 1.4: Convergence of bounded monotonic sequences. Every bounded, monotonic sequence converges. Calc 2 Lecture Notes Section 8.3 Page 3 of 5</p><p>For a Series Get a “feel” for the series: 1. Evaluate a few partial sums. 2. Graph the first few partial sums. 3. Evaluate some large-n partial sums.</p><p>If the terms are always positive: 1. Quick check for convergence: can you work out the answer to the sum? k 1 10 a. Geometric series: example: (0.1) = = (Section 8.2) k =0 1- 0.1 9 a ar k converges to if r <1and diverges if r 1 k =0 1- r b. Telescoping series: ゥ 1 1 1骣 1 1 骣 1 1 骣 1 1 邋 = - =琪 - + 琪 - + 琪 - + = 1 (Section 8.2) k=1k( k+1) k = 1 k k + 1桫 1 2 桫 2 3 桫 3 4 2. Quick check for convergence: is the series a p-Series? 1 a. The p-Series p converges if p > 1 and diverges if p  1. (Section 8.3) i=1 k 3. Quick check for divergence: do the terms grow instead of tending to zero?</p><p> k = limak 0 a.  divergence by the kth term test. Example: because k . k =1 (Section 8.2) 1 b. The harmonic series diverges: = . (Section 8.2) k =1 k 4. Develop an intuition for whether it will converge or diverge, then compare the sequence to a known converging or diverging integral or series: a. Theorem 3.1: The Integral Test for Convergence of a Series If f(k) = ak for all k = 1, 2, 3, …, and f is both continuous and decreasing, and f(x)  0 for x  1, then</p><p> f( x) dx and ak either both converge or both diverge. 1 i=1 b. Theorem 3.3: The Comparison Test for Convergence of a Series Suppose that </p><p>0  ak  bk for all k . If bk converges, then ak converges, too. If ak k =1 k =1 k =1</p><p> diverges, then bk diverges, too. k =1 c. Theorem 3.4: The Limit Comparison Test for Convergence of a Series a Suppose that a , b > 0, and that for some finite number L, limk =L > 0 . Then, k k k bk</p><p> either ak and bk both converge or both diverge. k =1 k =1 Calc 2 Lecture Notes Section 8.3 Page 4 of 5</p><p>If the terms of the series are not always positive, including if they alternate sign: 1. If the terms do go to zero, and they alternate sign, then the series converges. (Section 8.4)</p><p>2. If some of the terms are negative, but ak converges (which you would test using k =1 techniques from the previous page), then the series converges absolutely (Section 8.5)</p><p>3. Theorem 5.2: The Ratio Test Given ak , with ak  0 for all k, suppose that k =1 a lim k +1 = L . Then: k ak a. if L < 1, the series converges absolutely. b. if L > 1 (or L = ), the series diverges. c. if L =1, no conclusion can be made.</p><p>4. Theorem 5.3: The Root Test Given ak , with ak  0 for all k, suppose that k =1</p><p> k lim ak = L . Then: k a. if L < 1, the series converges absolutely. b. if L > 1 (or L = ), the series diverges. c. if L =1, no conclusion can be made. Calc 2 Lecture Notes Section 8.3 Page 5 of 5</p><p>Test When to Use Conclusions Section Geometric a 8.2 k r <1 Series ar Converges to if and k =0 1- r diverges if r 1.</p><p>Kth-Term Test All series. limak 0 8.2 If k , the series diverges. Integral Test 8.3 ak , where ak = f( k ) , f ak and f( x) dx both k =1 k =1 1 is continuous and converge or both diverge. decreasing, and f(x)  0 p-Series 1 Converges if p > 1 and diverges if 8.3 p p  1. k=1 k Comparison 0  a  b for all k . 8.3 k k b a Test If k converges, then k k =1 k =1 converges. </p><p>If ak diverges, then bk k =1 k =1 diverges.</p><p>Limit ak, bk > 0, and 8.3 a b Comparison a k and k both converge or limk =L > 0 k =1 k =1 Test k bk both diverge.</p><p>Alternating k +1 If limak = 0 and a< a , then the 8.4 -1 a k k+1 k Series Test ( ) k , where k =1 series converges.</p><p> ak > 0 Absolute Series with some positive 8.5 a a Convergence and some negative terms If k converges, then k (including alternating k =1 k =1 converges absolutely. series) Ratio Test Any series (especially 8.5 ak +1 those with exponentials or If lim = L and if L < 1, the k a factorials) k series converges absolutely.</p><p>Root Test Any series (especially k 8.5 If lim ak = L and if L < 1, the those with exponentials) k series converges absolutely.</p>