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MATEMÁTICAS 1215, PARTIAL 3 SOLUTIONS JOHN GOODRICK 1. Write down an example of a sequence which is bounded (acotada), but does not converge to any limit. SOLUTION: One example is the sequence an = (−1)n (i.e. the sequence −1, 1, −1, 1, . . .). The limit lim an does not exist, but this sequence is bounded n→∞ above (all the terms are less than 2) and bounded below (all the terms are greater than −2), so it is bounded. 2. For each of the following series, determine whether it is convergent or divergent, citing relevat test(s) for convergence. n ∞ 2 X n +1 (a) 2n2 − 3 n=1 SOLUTION: We use the Root Test: s n n2 + 1 n2 + 1 n lim = lim = 1/2. n→∞ n→∞ 2n2 − 3 2n2 − 3 Since this limit is less than 1, the series converges (absolutely). (b) ∞ X (−1)n n2 + 1 n=1 SOLUTION: This is an alternating series. The function f (x) = x21+1 is continuous, and limx→∞ x21+1 = 0. To check that f (x) is decreasing on the interval [1, ∞), note that 2x f 0 (x) = − 2 , (x + 1)2 and when x is positive, both 2x and (x2 + 1)2 are positive, so f 0 (x) < 0 when x > 0. Therefore, by the Alternating Series Test, this series converges. (c) ∞ X n=2 n (ln(n))2 1 2 JOHN GOODRICK SOLUTION: As n goes to infinity, the terms of the series (by L’Hopital’s Rule) tend to n n 1 n 1 = lim = lim = lim = ∞. = lim 2 n→∞ (ln(n)) n→∞ 2 ln(n) n→∞ 2 · (1/n) n→∞ 2 n→∞ 2 ln(n)1/n Since the limit of the terms is not zero, by the Test for Divergence, this series diverges. lim 3. (a) Show that the series ∞ X 2 ne−n converges. n=1 2 SOLUTION: Let f (x) = xe−x . The function f is continuous, and f (x) > 0 2 if x > 0 (since both x and e−x are positive, and so their product f (x) is positive). Also, the function f (x) is decreasing on [1, ∞) because 2 2 2 f 0 (x) = e−x − 2x2 e−x = e−x (1 − 2x2 ), 2 and if x ≥ 1 then 2x2 ≥ 2, so 1 − 2x2 < 0, and also e−x > 0 (since e raised to any power is positive), so their product f 0 (x) is negative. So by the Integral Test, the series converges if and only if the integral Z ∞ 2 xe−x dx 1 converges. Using the substitution u = x2 , du = 2 dx, the integral is Z t Z t2 1 −u −x2 xe e du lim dx = lim t→∞ 1 t→∞ 1 2 1 −t2 1 −1 1 = lim − e + e = . t→∞ 2 2 2e Since this limit exists (and is finite), the series converges. (b) Find a number B such that ∞ 3 X X 2 2 −n −n ne − ne ≤ B. n=1 n=1 The quantity within the absolute value signs is the remainder R3 for the sum. By the remainder estimate for the Integral Test (see section 11.3 of the textbook), Z ∞ Z t Z t2 1 −u −x2 −x2 |R3 | ≤ xe dx = lim xe dx = lim e du t→∞ 3 t→∞ 9 2 3 1 −t2 1 −9 1 = lim − e + e = 9. t→∞ 2 2 2e MATEMÁTICAS 1215, PARTIAL 3 So we can let B = SOLUTIONS 3 1 . 2e9 ∞ X 4n 4. Find the value of the sum . 5n+1 n=2 SOLUTION: The first few terms of this series are 42 43 44 42 4 42 + + + + . . . = (1 + + . . .). 53 54 55 53 5 52 The series inside the parentheses is a geometric series, with r = 54 . So the value of this series is 42 1 16 42 · ·5= . = 4 3 3 5 1− 5 5 25 5. Let f (x) = ln(1 + 2x) (considered as a function from the real numbers into the real numbers). (a) Find the first four terms of the Maclaurin series of f . SOLUTION: Recall that the nth term of the Maclaurin series is The first three derivatives of f are: f 0 (x) = f 00 (x) = − f (n) (0) n x . n! 2 = 2(1 + 2x)−1 1 + 2x 4 = −4(1 + 2x)−2 2 (1 + 2x) 16 = 16(1 + 2x)−3 (1 + 2x)3 So the first four terms of the Maclaurin series are f (3) (x) = f (0) + f 0 (0)x + f 00 (0) 2 f (3) (0) 3 8 x + x = 0 + 2x − 2x2 + x3 . 2 6 3 (b) Find the radius of convergence of the Maclaurin series of f . SOLUTION: If n ≥ 1, then continuing the pattern from part (a), we see that the nth derivative of f is f (n) (x) = (−1)n (n − 1)!2n (1 + 2x)−n . So the nth term of the Maclaurin series is 2n f (n) (0) n x = (−1)n xn . n! n 4 JOHN GOODRICK We use the Ratio Test to find values of x for which the Maclaurin series converges: n+1 n 2 n+1 2n|x| n+1 n n+1 x · (−1) n n = lim = 2|x| lim = 2|x|. lim (−1) n→∞ n→∞ n→∞ n + 1 n+1 2 x n Since the Maclaurin series converges when this limit is less than 1 and diverges when it is greater than 1, we see that the series converges for x in the interval (−1/2, 1/2) and diverges when x < −1/2 or when x > 1/2. Therefore the radius of convergence is 1/2. 6. Find all values of x for which the following series converges: ∞ X 3n xn √ . 3 n n=1 (HINT: It is not enough to find the radius of convergence. You must also test whether the series converges at the endpoints of the radius of convergence.) SOLUTION: First we find the radius of convergence using the Ratio Test: n+1 n+1 √ √ 3 3 3 x n 3|x| n = lim √ lim √ · 3 3 n n n→∞ n→∞ n+1 3 x n+1 r r n n = 3|x| 3 lim = 3|x|. = 3|x| lim 3 n→∞ n→∞ n + 1 n+1 Therefore the radius of convergence is 1/3. Next, we need to check whether the series converges when x = ±1/3. If x = −1/3, then the series is alternating, and 3n (−1/3)n (−1)n √ √ lim = lim = 0. 3 3 n→∞ n→∞ n n 1 Also, the function f (x) = √ 3 x is continuous and decreasing on [1, ∞), so by the Alternating Series Test, this series converges. If x = 1/3, then the series becomes ∞ X 1 √ . 3 n n=1 ∞ X 1 If n ≥ 1, then n ≤ n, so ≥ Since the harmonic series diverges, n n=1 by the Comparison Test, this series diverges, too. In conclusion, the series converges when x is in the interval [−1/3, 1/3) and diverges otherwise. √ 3 1 √ 3n 1 . n