Interval of convergence for series

Question

Find a power series for the function, centered at c, and determine the interval of convergence.

f(x) = 4/(5-x), c = -3

Answer 1

You can use the Taylor Series formula f(-3) + f'(-3)(x-(-3)) + f"(-3)(x-(-3))²/2! + f'''(-3)(x - (-3))³/3! + ... or

use the fact that 1 / (1-x) = 1 + x + x² + x³ + ..., where -1 < x < 1.

In the function f(x) = 4 / (5-x), replace x by x-(-3) and then manipulate the result to put it in the form

a(1 / (1-r)): 4 / (5-x) = 4 / (5 - (x+3) + 3) = 4 / (8 - (x+3)) = (4/8) [ 1 / (1 - (x+3)/8) ].

Next, replace x by (x+3)/8 in the formula 1/(1-x) = 1 + x + x² + ... to get:

= (1/2) [ 1 + (x+3)/8 + ((x+3)/8)² + ((x+3)/8)³ + ...], where -1 < (x+3)/8 < 1 (that is -11 < x < 5)

= ∑_{(n=0 to ∞)} [(x+3)ⁿ / (2(8ⁿ))], the interval of convergence is (-11, 5).