How do you find the power series for #f(x)=1/(1+3x)# and determine its radius of convergence?

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Answer 1 · 2017-01-09T17:24:04

#1/(1+3x) = sum_(n=0)^oo (-3x)^n# for #abs(x) < 1/3#

We know that the geometric series:

#sum_(n=0)^oo t^n#

converges for #abs(t) < 1# and the sum is:

#sum_(n=0)^oo t^n = 1/(1-t)#

We can then substitute:

#t=-3x#

and see that:

#1/(1+3x) = sum_(n=0)^oo (-3x)^n#

for

#abs (-3x) < 1#, or #abs(x) < 1/3#