4/(x + 4) - 3/(x + 1)

= 4/(4(¼x + 1)) - 3(x + 1)-¹

= 1/(1 + ¼x) - 3(1 + x)-¹

= (1 + ¼x)-¹ - 3(1 + x)-¹


Using Maclaurin expansion,


= (1 + (-1)(¼x) + (-1)(-2)/2! (¼x)² + ...) - 3(1 + (-1)x + (-1)(-2)/2! x² + ...)

= (1 - ¼x + 1/16 x² + ...) - 3(1 - x + x² + ...)

= 1 - ¼x + 1/16 x² - 3 + 3x - 3x² + ...

= -2 + 11/4 x - 47/16 x² + ...

Term in xⁿ

= (-1)(-2)...(-1 - n + 1) /n! (¼x)ⁿ - 3 [ (-1)(-2)...(-1 - n + 1)/n! xⁿ]

= (-1)(-2)...(-n) /n! (¼)ⁿxⁿ - 3[ (-1)(-2)...(-n)/n! xⁿ]

= (-1)(1)(-1)(2)...(-1)(n) /n! (¼)ⁿxⁿ - 3[ (-1)(1)(-1)(2)...(-1)(n) /n! xⁿ]

= (-1)ⁿn! /n! (¼)ⁿxⁿ - 3[ (-1)ⁿn! /n! xⁿ]

= (-1)ⁿ (¼)ⁿxⁿ - 3 (-1)ⁿxⁿ

= (-1)ⁿ xⁿ ( (¼)ⁿ - 3)

= (-x)ⁿ ( (¼)ⁿ - 3)

The left expansion is valid for |x| < 1

-1 < x < 1

The right expansion is valid for|¼x|<1

-1 < ¼x < 1
-4 < x < 4


For both to be valid overall,

-1 < x < 1

Range = (-1,1)