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

Question

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)

3247 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-10T10:20:56

See below.

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#