1. 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

2. 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

3. 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

4. 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

5. 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).