Is the solution correct? Question: 8, E(-1)" (n + 1) x" (converges for -1 < x < 1) to find a series that converges to (for -1 < x < 1). Use the fact that %3D (1+2)* (1+2) "Solution": Line 1: Notice that if f (x) = y then f' (x) (1+x)* (1+2) Line 2: Therefore, to find a series for -2 we could differentiate the series for (1+2)3 (1+a) Line 3: To find the derivative of a convergent series, we can differentiate term-by-term (i.e. differentiate one term at a time, as if we were differentiating a polynomial). Line 4: Let's expand the given series: = 1- 2x + 3x2 – 4x3 + 5x4 (1+z) 6x5 +... Line 5: Line 6: Now differentiate both sides: Line 7: (1+2) = -2 + 3. 2 – 4. 3x2 + 5 . 4x3 – 6· 5x4 +... Line 8: Write using sigma notation: Line 9: = E (-1)" (n + 1) · n. x" (1+z) n=0 Line 10: The radius of convergence does not change when differentiating, so E (-1)" (n + 1) · n. x" is a power series that converges to -2 for -1