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

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Answer 1 · 2017-12-09T15:49:49

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

converging for #absx < 1#.

Start the sum of the geometric series:

#sum_(n=0)^oo q^n = 1/(1-q)# for #absq < 1#

posing: #q = -x^2#, we have:

#sum_(n=0)^oo (-x^2)^n = sum_(n=0)^oo (-1)^n x^(2n) = 1/(1+x^2)# for #absx < 1#

In the interval of convergence we can integrate term by term:

#int dx/(1+x^2) = sum_(n=0)^oo int (-1)^n x^(2n) = sum_(n=0)^oo (-1)^n x^(2n+1)/(2n+1) + C#

As:

#int dx/(1+x^2) = arctanx +C#

we have:

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

as #arctan0 = 0# we can see that #C=0#, and dividing by #x# term by term:

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

converging for #absx < 1#.