How do you find the maclaurin series expansion of #f(x)= x / (1-x^4)#?

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Answer 1 · 2015-10-24T18:31:39

Use the Maclaurin series for #1/(1-t)# and substitution to find:

#x/(1-x^4) = sum_(n=0)^oo x^(4n+1)#

The Maclaurin series for #1/(1-t)# is #sum_(n=0)^oo t^n#

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

Substitute #t = x^4# to find:

#1/(1-x^4) = sum_(n=0)^oo x^(4n)#

Multiply by #x# to get:

#x/(1-x^4) = sum_(n=0)^oo x^(4n+1)#

This is a geometric series with common ratio #x^4#, hence the series converges when #abs(x) < 1# and the radius of convergence is #1#.