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Practice for Exam 2 - Math 3410 Intro to Analysis - Spring 2016 1. Determine whether each of the following series is convergent (absolutely or conditionally) or divergent. Indicate which test(s) you are using, and show all your work. ∞ ∞ p X X (a) ne−n (b) ( n2 + n − n) n=0 n=1 ∞ p X (c) ( n2 + 1 − n) (e) n=1 ∞ X n=1 sin n n2 + 1 (d) (f) ∞ X n! 1 · 3 · 5 · . . . · (2n − 1) n=1 ∞ X cos nπ n n=1 (g) ∞ X log n=1 n+1 n 2. Determine whether the series below is convergent or not 3 3 3 3 3 − + − + + ... 2 4 8 16 32 If convergent, find its sum. 3. Prove that lim sn = 3 where the sequence (sn ) is defined by sn = 1 3 5 2n − 1 + + 3 + ··· + , 2 22 2 2n n = 1, 2, 3, . . . . (Hint: see Problem 14.13(c)). 4. Show that ∞ X log n diverges when 0 < p < 1 and converges when p > 1. What happens when p = 1? np n=1 5. Prove that the monotonicity assumption in the Alternating Series Test is essential. That is, show that there exists P an alternating series (−1)n an , for which its sequence of positive terms (an ) satisfies lim an = 0, but which is a divergent series. P P an converges and bn converges absolutely, then an bn converges absolutely. P P P (b) Construct two convergent series an and bn for which the series an bn diverges. P P√ (c) Suppose an is a convergent series of positive terms. Show that the series an /n is also convergent. 6. (a) Show that if P P 7. Suppose an alternating series (−1)n+1 an , an > 0 and (an ) monotone, converges to a sum s. Show that the remainder Rn = s − sn satisfies |Rn | ≤ an+1 where sn is the nth partial sum of the series. 8. Suppose a series ∞ X an is convergent. Show that for every ε > 0, there exists N such that the remainder n=1 ∞ X an ≤ ε n=N +1 9. Define 1 1 1 + + · · · + − ln n . 2 3 n Prove that an is a decreasing sequence and that 0 < an < 1, hence the limit exists and is a number between 0 and 1. (Hint: See Problem 16.9) an = 1 + 10. [18 pts] Determine the radius of convergence and the interval of convergence for each of the power series (a) ∞ X n=1 n2 xn (b) ∞ X (−1)n 2n x n n=1 (c) ∞ X (4x − 3)n n2 2n n=1