We’ll need to use the ratio test. Part (a) We’ll need to use the ratio test. x lim x∞ (-1)n+1 (n+1) xn+1 (n+2) (n+1) (-1)n n xn 1 1 = x < 1 1 1 1 Now we need to test the endpoints. -1 < x < 1

Now we need to test the endpoints. -1 < x < 1 Part (a) If x = -1, then we have: 1/2 + 2/3 + 3/4 + 4/5 + … This series is approaching 1, so it diverges due to the nth-Term Test. If x = 1, then we have: -1/2 + 2/3 – 3/4 + 4/5 - … As x increases, this series begins to oscillate between -1 and 1, so it also diverges due to the nth-Term Test. Since both endpoints diverge, our official interval of convergence is -1 < x < 1. Now we need to test the endpoints. -1 < x < 1

Part (b) f’(x) = -1/2 + (4/3) x – (9/4) x2 + (16/5) x3 - … g’(x) = -1/2 + (1/12) x – (1/240) x2 + … g’(0) = -1/2 + (1/12) 0 – (1/240) 02 + … -1/2 y’(0) = f’(0) – g’(0) = -1/2 – (-1/2) y’(0) = 0

Part (b) f’(x) = -1/2 + (4/3) x – (9/4) x2 + (16/5) x3 - … g’(x) = -1/2 + (1/12) x – (1/240) x2 + … g”(x) = 1/12 - (1/120) x + … g”(0) = 1/12 - (1/120) 0 + … 1/12 y”(0) = f”(0) – g”(0) = 4/3 – 1/12 y”(0) = 5/4

Part (b) y’(0) = 0 y”(0) = 5/4 Since the 1st derivative = 0, x=0 must be a critical point. Since the 2nd derivative is positive at the same spot, x=0 is a relative minimum. y has a relative minimum at the point (0, -1).