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BC 3 Sequences and Series 2 Name: ____________________ Definition: An infinite series is an expression of the form a k 1 k a1 a2 a3 Also called an infinite sum or simply a series, we are basically adding infinitely many numbers together. You’ve been doing this for some time now. Remember, 1 k 1 MI-4 & Geometric Series: 2 1 . 1 k 1 2 1 2 3 3 3 1 . Fourth grade: 0 .33333 10 100 1000 3 In calculus, we are more thoughtful and define this idea carefully and logically. Some important terminology: Given the infinite series a k 1 k a1 a2 a3 , we define: n The nth partial sum, Sn by: Sn ak a1 a2 a3 k 1 The nth remainder, or nth tail by: Rn a k n 1 k an . an 1 an 2 an 3 The sequence of partial sums: S n n 1 . The sequence an n 1 is called the sequence of general terms. Note: i. a k 1 k S n Rn ii. Sn Sn1 an for n 1 Definition: We say the series a k 1 k converges if and only if the sequence of partial sums converges. In the case lim Sn S we say the infinite series converges to S and write n If a k 1 k does not converge, we say it diverges. IMSA BC3 Sequences and Series 2.1 a k 1 k S. Determine whether each of the following series converges by finding a simple formula for the partial sum S n and evaluating lim Sn . n (1) 1 (n 1)(n 1) n2 (2) k 1 (3) 1 3 k n 1 n 1 n e x dx This last problem suggests an intimate connection between infinite series and improper integrals, which we will explore more fully after a short digression into the theory of convergent and divergent series. IMSA BC3 Sequences and Series 2.2 Theorem (the nth term test): If lim an 0 , then n a n 1 n diverges. Proof: Left to the astute student. Determine (with explanation) whether the following series converge or diverge. If it converges, state the limit. (1) 1 1 n n 1 (2) 3 1 n n 1 n 1 4n (3) Find a formula for the nth partial sum of n ln n 1 . n 1 diverge? (4) 1 n (This is called the harmonic series) n 1 (5) Let’s look at the series 1 k!. k 0 IMSA BC3 Sequences and Series 2.3 Does the series converge or i. Explain why Sn is increasing. k 1 ii. Show that S n 1 for each n 0 . Does this imply Sn is bounded? k 0 2 n 1 iii. k n 1 k ! k 1 = _________________? k n 1 2 iv. Use your calculator or computer to find 1 k ! to the nearest thousandth. Explain k 0 carefully how you know your answer is within .001 One way to look at the previous problem is as follows: Suppose that an converges and n 1 n n 1 k 1 an S . Since S ak a k n 1 k S n Rn , we can say that Rn is the error when using S n to approximate S. We write S Sn Rn . So we often want to find a bound on Rn . IMSA BC3 Sequences and Series 2.4