How do I find the sum of a Taylor series sum_(n=0)^oo((-1)^n(x-2)^n)/(2^(n+1))?

How do I find the value of f(x)=sum_(n=0)^oo((-1)^n(x-2)^n)/(2^(n+1)). The interval of convergence is (0,4)

1 Answer
Apr 10, 2018

See below.

Explanation:

For abs y < 1 we have

sum_(k=0)^oo (-1)^ky^k = 1/(1+y)

considering now

y = (x-2)/2

we have

sum_(n=0)^oo((-1)^n(x-2)^n)/(2^(n+1)) = 1/2(1/(1+(x-2)/2))

for abs (x-2)/2 < 1