How do you find a power series representation for #f(x)= x/(9+x^2)# and what is the radius of convergence?

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Answer 1 · 2018-06-27T16:28:13

#x/(9+x^2) = sum_(k=0)^oo (-1)^k x^(2k+1)/3^(2k+2) #

with radius of convergence #R=3#.

Note that:

#x/(9+x^2) = x/9 1/(1+(x/3)^2)#

Consider the sum of the geometric series:

#sum_(k=0)^oo q^k = 1/(1-q)#

converging for #abs q < 1#

Let #q=-(x/3)^2# then:

#1/(1+(x/3)^2) = sum_(k=0)^oo (-(x/3)^2)^k = sum_(k=0)^oo (-1)^k x^(2k)/3^(2k)#

converging for #(x/3)^2 < 1#, that is for #x in (-3,3)#.

Now:

#x/(9+x^2) = x/9sum_(k=0)^oo (-1)^k x^(2k)/3^(2k) = sum_(k=0)^oo (-1)^k x^(2k+1)/3^(2k+2) #

with radius of convergence #R=3#.