Calc 2 Lecture Notes Section 8.3 Page 1 of 5
Which test do I use for a sequence or series?
Are you evaluating a sequence or a series (infinite sum)? If a series, go to page 3. Otherwise…
For a Sequence Calculate a few terms Graph some terms
Definition 1.1: Convergence of a sequence and divergence of a sequence.
The set {an} converges to L if and only if given any number > 0 there is an integer N for n= n0
which an - L < e for every n > N. If there is no such number L, then we say the sequence diverges.
Theorem 1.1: Combinations of limits of sequences.
If {an} and {bn} both converge, then n= n0 n= n0
lim(an+ b n) = lim a n + lim b n (i). n n n
lim(an- b n) = lim a n - lim b n (ii). n n n
(iii). lim(an b n) = lim a n lim b n n( n )( n )
lim an 骣an n (iv). lim 琪 = (assuming limbn 0). n n 桫bnlim b n n
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
Theorem 1.3: Squeeze Theorem
Suppose that {an} and {bn} both converge to the limit L. If there is an integer n1 n0 such n= n0 n= n0
that all n n1 guarantees that an cn bn, then {cn} converges to L as well. n= n0
Corollary 1.1:
liman = 0 liman = 0 If n , then n as well. Calc 2 Lecture Notes Section 8.3 Page 2 of 5
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.
Key trick: To determine whether a series is monotonic, look at the ratio of successive terms.
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.
Theorem 1.4: Convergence of bounded monotonic sequences. Every bounded, monotonic sequence converges. Calc 2 Lecture Notes Section 8.3 Page 3 of 5
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.
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?
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
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
0 ak bk for all k . If bk converges, then ak converges, too. If ak k =1 k =1 k =1
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
either ak and bk both converge or both diverge. k =1 k =1 Calc 2 Lecture Notes Section 8.3 Page 4 of 5
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)
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)
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.
4. Theorem 5.3: The Root Test Given ak , with ak 0 for all k, suppose that k =1
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
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.
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.
If ak diverges, then bk k =1 k =1 diverges.
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.
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.
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.
Root Test Any series (especially k 8.5 If lim ak = L and if L < 1, the those with exponentials) k series converges absolutely.