How Do You Solve This Maclaurin Series

The Maclaurin series for f(x) is given by:

1/2! - x^2/4! + x^4/6! - x^8/8! + …+ (-1)^n (x^2n)/ (2n+2)!+ …

a) For what values x does the given series converge?

b) Let g ‘ (x)= 1- x^2 * f(x). Write the Maclaurin series for g’(x), showing the first three nonzero terms and the general term.
c) Write g’(x) in terms of a familiar function without using series. Then write f(x) in terms of same familiar function.
d) Given that g(0)=3, write g(x) in terms of a familiar function without series.

Your 4th term should be x^6/8! not x^8/8!.

a) it converges for all x. For any fixed x this follows by the ratio test, or the alternating series test.
b) multiply the series by x^2 term by term and subtract from 1. You get 1 - x^2/2! + x^4/4! - x^6/6! + … with general term (-1)^n x^(2n)/(2n)!
c) g’(x) = cos(x). f(x) = (1-cos(x)) / x^2
d) integrate cos(x), get sin(x) + C, at 0 this must equal 3, so g(x) = sin(x) + 3.