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May 12, 2006 Lecturer: Dr Martin Kurth Trinity Term 2006 Problem Sheet 3 – Solutions 1. (a) ∞ µ ¶n−1 X 1 n=2 2 = ∞ µ ¶n−1 X 1 2 n=1 = = 1 1− 1 −1 −1 1 2 (b) ¶n ∞ µ X 1 − 4 n=1 = − ¶n−1 ∞ µ 1 1X − 4 n=1 4 1 1 ¡ ¢ 4 1 − − 41 1 = − 5 = − (c) ∞ X 2n n=1 µ ¶n−1 1 4 = = 2 2 ∞ X 2n−1 n=1 ∞ µ X n=1 = 2 = 4 1 1− 1 2 µ ¶n−1 1 4 ¶n−1 1 2 (d) ∞ X n=1 n n−1 2 3 µ ¶n+1 1 8 = = = = 1 µ ¶n−1 ∞ 2 X n−1 n−1 1 2 3 64 n=1 8 µ ¶ ∞ n−1 1 X 3 32 n=1 4 1 1 32 1 − 1 8 3 4 2. (a) Direct comparison test. and as P∞ 1 1 n3 cos2 (n) 1 ≤ 3, n3 n 2 P∞ converges, so does n=1 cosn3(n) . (b) Direct comparison test again. and as 1 n=1 n P∞ 1 2 − sin n ≥ , n n P∞ n diverges, so does n=1 2−sin . n (c) Yoohoo! Integral test! Define a decreasing function f by f (x) = e−2x 4 and calculate ∞ Z ¯b 1 lim e−2x ¯1 8 b→∞ 1 , 8e2 f (x)dx = − 1 = and as the integral converges, so does the series. (d) Integral test again. First we follow the hint and calculate ¡ ¢ µ ¶ sinh x1 d 1 cosh =− . dx x x2 Now we define our decreasing function f , ¡ ¢ sinh x1 f (x) = x2 and calculate Z ∞ f (x)dx = − lim cosh b→∞ 1 = cosh(1) − 1, µ ¶¯b 1 ¯¯ x ¯1 and as the integral converges, so does the series. 3. First we have to find the points where the graphs of f and g intersect. f (x) = g(x) 2 =⇒ x = x2 ie x = 0 or x = 1. Now we have to apply our washer cross section formula: Z 1³ ´ 2 2 V = π [f (x)] − [g(x)] dx 0 1 = π Z 1 = π Z = π µ 0 0 = ³ ¡ − 7 π 15 £ ¤2 ´ 2 [x + 1] − x2 + 1 dx ¢ −x4 − x2 + 2x dx ¶¯1 ¯ x5 x3 − + x2 ¯¯ 5 3 0 4. As the series is convergent, we must have lim an = 0. n→∞ Now split the sequence {an } into its P positive and negative subsequences ∞ {cn }, P cn ≥ 0 and {dP n }, dn < 0. As n=1 is not absolutely convergent, ∞ ∞ d must diverge. Now we can construct our c and both n n n=1 P∞ n=1 rearranged series b in the following way: Take exactly as many n n=1 terms from {cn } as you need to make the sum just greater than L. Then add terms from {dn }, such that the sum is (just) less than L. Continue this procedure until infinity. After each step, the difference ¯ ¯ n ¯X ¯ ¯ ¯ ²n = ¯ bk − L¯ ¯ ¯ k=1 is less than or equal to the absolute value of the last term included in the series. As limn→∞ an = 0, we have limn→∞ ²n = 0, and thus ∞ X n=1 3 bn = L.